If the mean of P, Q,R is A and PQ + QR + RP =0, then the mean of p^2,Q^2,R^2 is. A) 2A^2 B) 4A^2 C) A^2 D) 3A^2

If the mean of P, Q,R is A and PQ + QR + RP =0, then the mean of p^2,Q^2,R^2 is. A) 2A^2 B) 4A^2 C) A^2 D) 3A^2
by

1 Answer

Answer :

D)

We have (P + Q + R)/3 = A

P+Q+R = 3A

(P+Q+R)^2 = 9A^2

P^2+Q^2+R^2 + 2 (PQ + QR + PR) = 9A^2

P^2+Q^2+R^2 = 9A^2

Required mean = (P^2+Q^2+R^2)/3 = 9A^2/3= 3A^2
by

Related questions

If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Description : If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Last Answer : Solution :-

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

The average weight of M,N and O is 168 kg. If P joins the group, the average weight of the group becomes 160 kg. If another man Q who weighs 6kg more than P replaces M, then the average of N,O,P and Qbecomes158 kg. What is the weight of M? A) 175 B) 140 C) 145 D) 150

Description : The average weight of M,N and O is 168 kg. If P joins the group, the average weight of the group becomes 160 kg. If another man Q who weighs 6kg more than P replaces M, then the average of N,O,P and Qbecomes158 kg. What is the weight of M? A) 175 B) 140 C) 145 D) 150 

Last Answer : D) M+N+O =168 * 3 =504 M+N+O+P=160*4=640……(1) P=136 kg and Q =142 kg N+O+P+Q =158*4 =632 N+O+P =632-142 =490…..(2) from 1 & 2, M=640-490 =150kg

Distance between two stations P & Q is 1556km. A passenger train covers the journey from P to Q at 168km per hr and return back to P with a uniform speed of 112km/hr. Find the Average speed of the train during the whole journey? A) 124.4 km/hr B) 130.4km/hr C) 134.4 km/hr D) 130.0 km/hr

Description : Distance between two stations P & Q is 1556km. A passenger train covers the journey from P to Q at 168km per hr and return back to P with a uniform speed of 112km/hr. Find the Average speed of the train during the whole journey? A) 124.4 km/hr B) 130.4km/hr C) 134.4 km/hr D) 130.0 km/hr

Last Answer : C) Given x= 168 , y = 112 Required average speed = (2xy / x + y) km/hr = 2 * 168 *112 / 168 + 112 = 37632 / 280 =134.4 km/hr

`PQR` is a triangular park with `PQ=PR=200m`. A.T.V. tower stands at the mid-point of `QR`. If the angles of elevation of the top of the tower at `P`,

Description : `PQR` is a triangular park with `PQ=PR=200m`. A.T.V. tower stands at the mid-point of `QR`. If the angles of elevation of the top of the tower at `P`,

Last Answer : `PQR` is a triangular park with `PQ=PR=200m`. A.T.V. tower stands at the mid-point of `QR`. If the angles of ... (3)` B. `50sqrt(2)` C. `100` D. `50`

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Description : If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Last Answer : (b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW. Here, given AB = ... , or ΔCBA ≅ ΔPRQ, so option (b) is correct, or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Description : If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Last Answer : (b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW. Here, given AB = ... , or ΔCBA ≅ ΔPRQ, so option (b) is correct, or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Description : If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Last Answer : Solution :- △ ABC ≅ △PQR

O is a point on side PQ of a APQR such that PO = QO = RO, then (a) RS2 = PR × QR (b) PR2 + QR2 = PQ2 (c) QR2 = QO2 + RO2 (d) PO2 + RO2 = PR2

Description : O is a point on side PQ of a APQR such that PO = QO = RO, then (a) RS2 = PR × QR (b) PR2 + QR2 = PQ2 (c) QR2 = QO2 + RO2 (d) PO2 + RO2 = PR2

Last Answer : (b) PR2 + QR2 = PQ2

Prove that (sin A + sin 3A + sin 5A + sin7 A)/(Cos A + Cos3 A + cos5 A + cos7 A) = tan 4A. -Maths 9th

Description : Prove that (sin A + sin 3A + sin 5A + sin7 A)/(Cos A + Cos3 A + cos5 A + cos7 A) = tan 4A. -Maths 9th

Last Answer : answer:

4a - a x 3a + 7a?

Description : 4a - a x 3a + 7a?

Last Answer : 30

Which is the latest satellite of India placed in the geosynchronous orbit ? (1) INSAT-2D (2) INSAT-3A (3) INSAT-4A (4) Kalpana

Description : Which is the latest satellite of India placed in the geosynchronous orbit ? (1) INSAT-2D (2) INSAT-3A (3) INSAT-4A (4) Kalpana

Last Answer : INSAT-3A

If ` a : b =2 : 3 ," then find "(3a - b) : ( 2a+ 3b)`.

Description : If ` a : b =2 : 3 ," then find "(3a - b) : ( 2a+ 3b)`.

Last Answer : If ` a : b =2 : 3 ," then find "(3a - b) : ( 2a+ 3b)`.

Find the sum of the value (s) of a so that the equation (`x^(2) + 2ax + 2a + 3`) (`x^(2) + 2ax + 4a +5 `) = 0 has only 3 real distinct roots.

Description : Find the sum of the value (s) of a so that the equation (`x^(2) + 2ax + 2a + 3`) (`x^(2) + 2ax + 4a +5 `) = 0 has only 3 real distinct roots.

Last Answer : Find the sum of the value (s) of a so that the equation (`x^(2) + 2ax + 2a + 3`) (`x^(2) + 2ax + 4a +5 `) = 0 has only 3 real distinct roots.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

In angle PQR angle P = 80 .If PQ =PR find angle Q and angle R -Maths 9th

Description : In angle PQR angle P = 80 .If PQ =PR find angle Q and angle R -Maths 9th

Last Answer : a triangle includes 3 angles summing up as 180° so 180 =

If pq plus qr pr statements must be true?

Description : If pq plus qr pr statements must be true?

Last Answer : What is the answer ?

The diagram below shows the propagations of a wave. Which points are in phase? (a) PQ (b) QR (c) QS (d) TQ

Description : The diagram below shows the propagations of a wave. Which points are in phase? (a) PQ (b) QR (c) QS (d) TQ

Last Answer : Ans:(c)

An independent voltage source in series with an impedance ZS=R+jXS delivers a maximum average power to

Description :  An independent voltage source in series with an impedance ZS=R+jXS delivers a maximum average power to a load impedance ZL when A) ZL = R B) ZL = jXS C) ZL = R-jXS D) ZL = R+jXS 

Last Answer :  An independent voltage source in series with an impedance ZS=R+jXS delivers a maximum average power to a load impedance ZL when ZL = R-jXS

What is 3x plus 2a plus 5x plus 4a?

Description : What is 3x plus 2a plus 5x plus 4a?

Last Answer : 58

In a simply supported beam (l + 2a) with equal overhangs (a) and carrying a uniformly distributed load over its entire length, B.M. at the middle point of the beam will be zero if (A) l = 2a (B) l = 4a (C) l < 2a (D) l > a

Description : In a simply supported beam (l + 2a) with equal overhangs (a) and carrying a uniformly distributed load over its entire length, B.M. at the middle point of the beam will be zero if (A) l = 2a (B) l = 4a (C) l < 2a (D) l > a

Last Answer : (A) l = 2a

A simply supported beam (l + 2a) with equal overhangs (a) carries a uniformly distributed load over the whole length, the B.M. changes sign if (A) l > 2a (B) l < 2a (C) l = 2a (D) l = 4a

Description : A simply supported beam (l + 2a) with equal overhangs (a) carries a uniformly distributed load  over the whole length, the B.M. changes sign if  (A) l > 2a (B) l < 2a (C) l = 2a (D) l = 4a

Last Answer : (A) l > 2a

A ship sails out to a mark at the rate of 10 km per hour and sails back at th rate of 15 km per hour. What is its average rate of sailing? (a) 10 km. p.h. (b) 12 km. p.h. (c) 15 km. p.h. (d) 11 km. p.h.

Description : A ship sails out to a mark at the rate of 10 km per hour and sails back at th rate of 15 km per hour. What is its average rate of sailing? (a) 10 km. p.h. (b) 12 km. p.h. (c) 15 km. p.h. (d) 11 km. p.h.

Last Answer : 12 km. p.h.

A man wnt uphill with a speed of 20 km.p.h. and came downhill with a speed of 30 km p.h. The average speed for his journey was (a) 25 km. p.h. (b) 22 1/2 km. p.h. (c) 24 km. p.h. (d) 25 1/2 km. p.h.

Description : A man wnt uphill with a speed of 20 km.p.h. and came downhill with a speed of 30 km p.h. The average speed for his journey was (a) 25 km. p.h. (b) 22 1/2 km. p.h. (c) 24 km. p.h. (d) 25 1/2 km. p.h.

Last Answer : 24 km.p.h.

What is 3a-1-a-2a plus 3?

Description : What is 3a-1-a-2a plus 3?

Last Answer : It is then 2 when all terms are added together

The pair of equations ax+2y=7 and 3x+by=16 represent parallel lines if (a)a=b (b)3a=2b (c)ab=6 (d)2a=3b

Description : The pair of equations ax+2y=7 and 3x+by=16 represent parallel lines if (a)a=b (b)3a=2b (c)ab=6 (d)2a=3b

Last Answer : (c)ab=6

The maximum magnitude of shear stress due to shear force F on a rectangular section of area A at the neutral axis, is (A) F/A (B) F/2A (C) 3F/2A (D) 2F/3A

Description : The maximum magnitude of shear stress due to shear force F on a rectangular section of area A at  the neutral axis, is  (A) F/A (B) F/2A (C) 3F/2A (D) 2F/3A

Last Answer : (C) 3F/2A

18 friends went to a restaurant for taking their breakfast. 17 of them spent RS.24 each on their breakfast and the last one spent RS.16 more then the average expenditure of all the 18. What was the total money spent by them? A) 228 B) 448.9 C) 458.6 D) 428.0

Description : 18 friends went to a restaurant for taking their breakfast. 17 of them spent RS.24 each on their breakfast and the last one spent RS.16 more then the average expenditure of all the 18. What was the total money spent by them? A) 228 B) 448.9 C) 458.6 D) 428.0 

Last Answer : B)  Let the average expenditure of all the 18 be Rs ‘x’.  Then (24 * 17) + (x + 16) = 18x 424 + x = 18x x = 24.9 therefore total money spent 18x  = 18 * 24.9  = Rs.448.9

If the arithmetic mean of 150 numbers is calculated 70. If each number is increased by 10, then mean of new number is. A) 70 B) 80 C) 90 D) 60

Description : If the arithmetic mean of 150 numbers is calculated 70. If each number is increased by 10, then mean of new number is. A) 70 B) 80 C) 90 D) 60

Last Answer : B) Arithmetic mean of numbers = 70 Sum of 150 numbers = (70 * 150) =10500 Total increase = (150 * 10) = 1500 Increased sum = 10500 + 1500 = 12000 Increased average = 12000/150 =80

Let us assume that you construct ordered tree to represent the compound proposition (~(p˄q))↔(~p˅~q). Then, the prefix expression and post-fix expression determined using this ordered tree are given as ........... and ............. respectively. (A) ↔~˄pq˅ ~~pq, pq˄~p~q~˅↔ (B) ↔~˄pq˅ ~p~q, pq˄~p~q~˅↔ (C) ↔~˄pq˅ ~~pq, pq˄~p~ ~q˅↔ (D) ↔~˄pq˅ ~p~q, pq˄~p~~q˅↔

Description : Let us assume that you construct ordered tree to represent the compound proposition (~(p˄q))↔(~p˅~q). Then, the prefix expression and post-fix expression determined using this ordered tree are given as ........... and .......... ... ~p~q~˅↔ (C) ↔~˄pq˅ ~~pq, pq˄~p~ ~q˅↔ (D) ↔~˄pq˅ ~p~q, pq˄~p~~q˅↔

Last Answer : (B) ↔~˄pq˅ ~p~q, pq˄~p~q~˅↔

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Last Answer : 1.Draw a line segment QR = 5 cm. 2.With Q as centre, construct an angle of 90° and let this line through Q is QX. 3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Last Answer : 1.Draw a line segment QR = 5 cm. 2.With Q as centre, construct an angle of 90° and let this line through Q is QX. 3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

on analyzing the result of an competitive exam the teacher found that the average for the entire the class was 69 marks. If we say that average of 10 % of the students scored 77 marks and average of 28 % of the students scored 66 marks, then calculate average marks of the remaining students of the class A) 67.54 B) 68.26 C) 66.91 D) 69.06

Description :  on analyzing the result of an competitive exam the teacher found that the average for the entire the class was 69 marks. If we say that average of 10 % of the students scored 77 marks and average of 28 % of the ... marks of the remaining students of the class A) 67.54 B) 68.26 C) 66.91 D) 69.06 

Last Answer : D) average of entire class = 69marks average of 28 % of the students = 66 marks average of 10% of the students = 77 marks then % of remaining students = (100- 10 - 28) = 62% let the average of 62% of the ... (62*x)+770+1848= 6900 62 * x = 6900-770-1848 62*x = 4282 x = 69.06

The average expenditure of A, Band C is Rs 10000 per month. Also, the average expenditure of B, C and D is Rs 14000 per month. If the average expenditure of D is thrice of that of A then the average expenditure of B and C is: A) 12000 B) 15000 C) 18000 D) 21000

Description : The average expenditure of A, Band C is Rs 10000 per month. Also, the average expenditure of B, C and D is Rs 14000 per month. If the average expenditure of D is thrice of that of A then the average expenditure of B and C is: A) 12000 B) 15000 C) 18000 D) 21000

Last Answer : A) Total expenditure of A, B & C=3×10000=30000 Total expenditure of B, C & D =3×14000=42000 D-A=42000-30000=12000 Also, we are given D=3A. so 3A-A =12000 & A=6000 Total expenditure of B & C=30000-6000=24000 Average expenditure of B& C=24000/2 =12000

Average age of A, B, C is 90yrs, however when D joins them, then the average comes down to 80 yrs. Now a new person E, whose age is six-fifth of the age of D, replaces A and the new average is 78. What is age of A? A) 64 B) 68 C) 62 D) 66

Description : Average age of A, B, C is 90yrs, however when D joins them, then the average comes down to 80 yrs. Now a new person E, whose age is six-fifth of the age of D, replaces A and the new average is 78. What is age of A? A) 64 B) 68 C) 62 D) 66

Last Answer : B) according to the question A+B+C/3=90 => A+B+C =270…………..(1) A+B+C +D/4=80 => A+B+C +D=320………(2) From the equation 1 and 2 D=50 E =6/5×50=60 E+B+C +D/4=78 => E+B+C +D =312 B+C +D =312-60=252……………..(3) From 2and 3 The age of A=320-252=68

If the average of two numbers are 138& product is 2241. Then find the difference of both numbers? A) 225.40 B) 259.25 C) 267.81 D) 294.57

Description : If the average of two numbers are 138& product is 2241. Then find the difference of both numbers? A) 225.40 B) 259.25 C) 267.81 D) 294.57

Last Answer : B) let the two number be a & b sum of two numbers, a+b = 138*2 =276---1 product of two numbers, a*b=2241 (a+b)2 = a2 + 2ab + b2 ------2 (a-b)2 = a2 -2ab +b2 ------3 solving 2 & 3, 76176 - (a-b)2 = 4*2241 a-b = 259.25

Students of two colleges appeared for Talent test carrying 250marks as maximum. The average of their marks for college1 & college 2 are 160 & 180respectively. If the number of students of college 1 is the half of the number of students of college 2, then what is the average marks of all students of both the college? A) 170 B) 110 C) 173.33 D) 177.33

Description : Students of two colleges appeared for Talent test carrying 250marks as maximum. The average of their marks for college1 & college 2 are 160 & 180respectively. If the number of students of college 1 is the half of the ... marks of all students of both the college? A) 170 B) 110 C) 173.33 D) 177.33

Last Answer : C) let the number of students of college 2 be '2N' then the number of students of college1 is 'N' the average marks for college1 is 160 the average marks for college 2 is 180 total marks of ... = 360N average marks of all students of both the colleges = (160N + 360N)/N+2N = 173.33marks

If the average of 4 observations x, x+1, x+2, x+3 is 12 then the average of last 2 observations is A) 10 B) 11 C) 12 D) 13If the average of 4 observations x, x+1, x+2, x+3 is 12 then the average of last 2 observations is A) 10 B) 11 C) 12 D) 13

Description : If the average of 4 observations x, x+1, x+2, x+3 is 12 then the average of last 2 observations is A) 10 B) 11 C) 12 D) 13

Last Answer :  D) We have (x+x+1+x+2x+3/40=12 4x+6=48 4x=42 x=10.5 so the numbers are 10.5,11.5, 12.5, 13.5 required average = 12.5+13.5/2 =26/2 =13.

An industry employed 1200 men and 800 women and the average salary was Rs 51 per day. If a woman got Rs 10 less than a man , then what are their daily salary?(mens and women respectively) A) men Rs 25 and women Rs 35 B) men Rs 45 and women Rs 55 C) men Rs 35 and women Rs 25 D) men Rs 55 and women Rs 45

Description : An industry employed 1200 men and 800 women and the average salary was Rs 51 per day. If a woman got Rs 10 less than a man , then what are their daily salary?(mens and women respectively) A) men Rs 25 and women Rs ... Rs 45 and women Rs 55 C) men Rs 35 and women Rs 25 D) men Rs 55 and women Rs 45

Last Answer : D) Let the daily salary of man be Rs’x’ Then the daily salary of a women = Rs (x-100 Now 1200x + 800(x-10) = 51 *(1200+800) 1200x+800x -8000 = 51*2000 2000x = 102000+8000 =110000 X=55 Therefore mens daily salary Rs 55 and womens daily salary Rs 45

The average state of the city in the first 4 days of a month was 116 degrees. The average for the second, third, fourth and fifth days was 120 degrees. If the state of the first and fifth days were in the ratio 7:8 then what is the state on the fifth day. A) 112degrees B) 128degrees C) 132degrees D) 130degrees

Description : The average state of the city in the first 4 days of a month was 116 degrees. The average for the second, third, fourth and fifth days was 120 degrees. If the state of the first and fifth days ... :8 then what is the state on the fifth day. A) 112degrees B) 128degrees C) 132degrees D) 130degrees

Last Answer :  B)  Sum of states on 1st, 2nd, 3rd, 4th days = (116 *4) = 464 =======>A Sum of states on 2nd, 3rd, 4th and 5th days = (120 * 4)= 480 ======>b From a and b Subtracting each other ... be 7x and 8x degrees respectively Then 8x - 7x = 16 x = 16 therefore state on the 5th day = 8x = 128 degrees.

The average age of 78 students of a batch is 30 years. If the age of the head master be included, then the average increases by 6 months. Find the age of head master? A) 56yrs 5 months B) 66yrs 6 months C) 56yrs 5 months D) 69yrs 5 months

Description : The average age of 78 students of a batch is 30 years. If the age of the head master be included, then the average increases by 6 months. Find the age of head master? A) 56yrs 5 months B) 66yrs 6 months C) 56yrs 5 months D) 69yrs 5 months

Last Answer : D)  Total age of 78 students = (780 * 30)yrs  = 2340 yrs  Average of 79 persons = 30 yrs 6 months  = 6 ½ yrs  Total age of 79 persons = 61/2 * 79  = 2409.5  Age of headmaster = (2409.5 – 2340)yrs  = 69.5 yrs = 69 yrs 5 months.

In an exam the average marks obtained by a candidate is 82 per paper. If he had obtained 32 marks more in science paper & 28more marks in social paper, then his average per paper is increased by 15 marks. Then how many papers when there in examination? A) 10 B) 6 C) 4 D) 8

Description : In an exam the average marks obtained by a candidate is 82 per paper. If he had obtained 32 marks more in science paper & 28more marks in social paper, then his average per paper is increased by 15 marks. Then how many papers when there in examination? A) 10 B) 6 C) 4 D) 8 

Last Answer : C)  let the number of paper be x total mark earned him = 82x then,, 82x+32+28= 97x 15x=60 x= 4 =number of subjects

The average monthly salary of directors & office boys of an organization Rs. 3750. Then the average salary of all directors is Rs. 6000 while that of all office boys is Rs. 3000 per month. If there are 120 employees in the organization then find the ratio of manager and office boy. A) 2:5 B) 3:1 C) 1:3 D) 5:2

Description : The average monthly salary of directors & office boys of an organization Rs. 3750. Then the average salary of all directors is Rs. 6000 while that of all office boys is Rs. 3000 per month. If there are 120 employees in the ... find the ratio of manager and office boy. A) 2:5 B) 3:1 C) 1:3 D) 5:2 

Last Answer :  C)  let the number of director be X and office boy be 120-x total salary of director and office boy = 120*3750 = Rs.450000 total salary of director = 6000x total salary of office boy = 3000* (120 - ... director number of office boy = 120-30 = 90 ratio of director to office boy = 30:90 = 1:3

There were 32 boys in a hostel. If the number of boys be increased by 15, then the expenditure on food increases by Rs. 43 per day while the average of expenditure of boys is reduced by Rs. 2. What was the initial expenditure on food per day? A) Rs.108.2 B) Rs.125.6 C) Rs.135.8 D) Rs.144.5

Description : There were 32 boys in a hostel. If the number of boys be increased by 15, then the expenditure on food increases by Rs. 43 per day while the average of expenditure of boys is reduced by Rs. 2. What was the initial expenditure on food per day? A) Rs.108.2 B) Rs.125.6 C) Rs.135.8 D) Rs.144.5

Last Answer : Answer: A)  Let initial expense of 32 boys = Rs. x average expense of 32 boys = x/32 average expense of 47 boys = (x + 43)/ 47 then, (x/32) - ((x+43)/47) = 2  x= Rs.108.28

Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5

Description : Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5 

Last Answer : Answer: B) Maximum total marks = 5 120 = 600 Marks scored by Saravanan = 90/100*600=540 Let the marks scored by him in the 5 subjects be 18x,14x,15x,21x and 22x respectively. 18x + 14x + 15x + 21x + ... A' grade i.e. 0.85 120 = 102 Therefore, he has got an A' grade in 3 subjects.

The average age of the assembalage of people was 114 yrs. When the 2 people of the assembalage went out with age of age of 84 yrs and 108yrs, then the average of the assembalage increased by 12yrs. How many people are there initially? A) 3 B) 4 C) 5 D) 6

Description : The average age of the assembalage of people was 114 yrs. When the 2 people of the assembalage went out with age of age of 84 yrs and 108yrs, then the average of the assembalage increased by 12yrs. How many people are there initially? A) 3 B) 4 C) 5 D) 6 

Last Answer : Answer: C)  let the number of people initially be N total age of N people = 114N total age of N-2 people = 114N - ( 84+108) = 114N -192 average age of N-2 people = 126 = [ 114N - 192] / [ N - 2] N=5 Hence th required answer is 5