If `x^(-(p/q))=(1/x)^(k)` then `k=` ________.

If `x^(-(p/q))=(1/x)^(k)` then `k=` ________.
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If `x^(-(p/q))=(1/x)^(k)` then `k=` ________.
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If `((x^(m)xx x^(n))/x^(p))^(q)=x^(k)`, then `k=` ________.

Description : If `((x^(m)xx x^(n))/x^(p))^(q)=x^(k)`, then `k=` ________.

Last Answer : If `((x^(m)xx x^(n))/x^(p))^(q)=x^(k)`, then `k=` ________.

If `a^(p/q)=root(q)(k)`, then `k=` _______.

Description : If `a^(p/q)=root(q)(k)`, then `k=` _______.

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Aniline reacts with mixed acid (conc. `HNO_(3)` and conc. `H_(2)SO_(4)`) at 288 K to give P (51%), Q (47%) and R(2%). The major product (s) of the fol

Description : Aniline reacts with mixed acid (conc. `HNO_(3)` and conc. `H_(2)SO_(4)`) at 288 K to give P (51%), Q (47%) and R(2%). The major product (s) of the fol

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If vectors P = 2i + 3j - k and Q = 2i - 5j + 2k. Find i. P + Q ii. 3P - 2Q

Description : If vectors P = 2i + 3j - k and Q = 2i - 5j + 2k. Find i. P + Q ii. 3P - 2Q

Last Answer : If vectors \(\overset\rightarrow{P}\) = 2i + 3j - k and \(\overset\rightarrow{Q}\) = 2i - ... \overset\rightarrow{P}\) - 2\(\overset\rightarrow{Q}\)

Find the potential between two points p(1,-1,0) and q(2,1,3) with E = 40xy i + 20x 2 j + 2 k a) 104 b) 105 c) 106 d) 107

Description : Find the potential between two points p(1,-1,0) and q(2,1,3) with E = 40xy i + 20x 2 j + 2 k a) 104 b) 105 c) 106 d) 107

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The formula Q = P - K [1.8T + 32] in which Q is runoff, P is annual rain fall in cm, T is mean annual temperature in centigrade and K is a constant, is known (A) Justin's formula (B) Khosla's formula (C) English formula (D) Vermule's formula

Description : The formula Q = P - K [1.8T + 32] in which Q is runoff, P is annual rain fall in cm, T is mean annual  temperature in centigrade and K is a constant, is known  (A) Justin's formula  (B) Khosla's formula  (C) English formula  (D) Vermule's formula

Last Answer : (B) Khosla's formula 

If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th

Description : If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th

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If `lim_(x rarr -3) (x^(k)+3^(k))/(x+3)=405`, where k is an odd natural number then, k = ________.

Description : If `lim_(x rarr -3) (x^(k)+3^(k))/(x+3)=405`, where k is an odd natural number then, k = ________.

Last Answer : If `lim_(x rarr -3) (x^(k)+3^(k))/(x+3)=405`, where k is an odd natural number then, k = ________.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Description : If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Last Answer : x3+px+q 3x2+p have common root Let common root =a a3+ap+q=0 a2=3−p​a=3−p​​4p3+27q2=? a3(a2+p)+q=0 ⇒3−p​​[(3−p​)+p]+q=0 3−p​​(3+2p​)=−q ⇒27−p(4p2)​=q2

The normal at a point `P` on the ellipse `x^2+4y^2=16` meets the x-axis at `Qdot` If `M` is the midpoint of the line segment `P Q ,` then the locus of

Description : The normal at a point `P` on the ellipse `x^2+4y^2=16` meets the x-axis at `Qdot` If `M` is the midpoint of the line segment `P Q ,` then the locus of

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If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

The inherent characteristic of an equal percentage valve relating flow rate 'q' with valve stem movement 'x' are described by the equation (A) dq/dx = K (B) dq/dx = K.q (C) dq/dx = K/q (D) dq/dx = Kq 2

Description : The inherent characteristic of an equal percentage valve relating flow rate 'q' with valve stem movement 'x' are described by the equation (A) dq/dx = K (B) dq/dx = K.q (C) dq/dx = K/q (D) dq/dx = Kq 2

Last Answer : (A) dq/dx = K

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If the coordinates of two points are P( -2,3) and Q ( -3, 5) then find (abscissa of P)–(abscissa of Q) -Maths 9th

Description : If the coordinates of two points are P( -2,3) and Q ( -3, 5) then find (abscissa of P)–(abscissa of Q) -Maths 9th

Last Answer : Abscissa of P – Abscissa of Q = (–2) – (–3) = –2 + 3 = 1.

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 :-

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Description : If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Last Answer : (a) 0Let h be the first term and k be the common ratio of a GP, then a = hkp - 1, b = hkq - 1, c = hkr - 1∴ (q - r) log a + (r - p) log b + (p - q) log c = log [hkp -1]q - r + log [hkq -1]r - p + log[hkr -1]p - ... r + r - p + p - q) (kp - 1)q - r (kq -1)r - p (kr -1)p - q = log(ho ko) = log 1 = 0.

For three distinct positive numbers p, q and r, if p + q + r = a, then -Maths 9th

Description : For three distinct positive numbers p, q and r, if p + q + r = a, then -Maths 9th

Last Answer : answer:

If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Description : If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Here, we are joining A and C. In ΔABC P is the mid point of AB Q is the mid point of BC PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and ... RS=PS=RQ[All sides are equal] ∴ PQRS is a parallelogram with all sides equal ∴ So PQRS is a rhombus.

ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Description : ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Last Answer : Here is your First of all we will draw a quadrilateral ABCD with AD = BC and join AC, BD, P,Q,R,S are the mid points of AB, AC, CD and BD respectively. In the triangle ABC, P and Q are mid points of AB and AC respectively. All sides are equal so PQRS is a Rhombus.

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

Let `P={theta:sintheta-costheta=sqrt2cos theta}and Q={theta:sintheta+costheta=sqrt2sintheta}` be two ses. Then,

Description : Let `P={theta:sintheta-costheta=sqrt2cos theta}and Q={theta:sintheta+costheta=sqrt2sintheta}` be two ses. Then,

Last Answer : Let `P={theta:sintheta-costheta=sqrt2cos theta}and Q={theta:sintheta+costheta=sqrt2sintheta}` be two ses. ... QcancelsubeP` C. `PcancelsubeQ` D. `P=Q`

The price of a refrigetor, P is 5 times that of a cooler Q. If then price of P increases by `40%.` then what is the change in the total price of the r

Description : The price of a refrigetor, P is 5 times that of a cooler Q. If then price of P increases by `40%.` then what is the change in the total price of the r

Last Answer : The price of a refrigetor, P is 5 times that of a cooler Q. If then price of P ... total price of the refrigerator and cooler put together ?

If `(p)/(q)=(r)/(s)=(t)/(u)` then `(p^(2)+r^(2)+t^(2))/(q^(2)+s^(2)+u^(2))=`

Description : If `(p)/(q)=(r)/(s)=(t)/(u)` then `(p^(2)+r^(2)+t^(2))/(q^(2)+s^(2)+u^(2))=`

Last Answer : If `(p)/(q)=(r)/(s)=(t)/(u)` then `(p^(2)+r^(2)+t^(2))/(q^(2)+s^(2)+u^(2))=` A. `(prt)/(qsu)` B. `1 ... q^(3)+s^(3)+u^(3))` D. `((p+r+t)/(q+s+u))^(2)`

If P:Q:R=2:3:4 and `P^(2)+Q^(2)+R^(2)=11600`, then find `(P+Q-R)`

Description : If P:Q:R=2:3:4 and `P^(2)+Q^(2)+R^(2)=11600`, then find `(P+Q-R)`

Last Answer : If P:Q:R=2:3:4 and `P^(2)+Q^(2)+R^(2)=11600`, then find `(P+Q-R)` A. 15 B. 16 C. 18 D. 20

If p and q are perfect squares then `sqrt(p)/(q)` is always a rational number .Is the statement true?

Description : If p and q are perfect squares then `sqrt(p)/(q)` is always a rational number .Is the statement true?

Last Answer : If p and q are perfect squares then `sqrt(p)/(q)` is always a rational number .Is the statement ... B. no C. cannot be determined D. none of these

If `(2p+5q)/(2r+5s) = (4p-3q)/(4r-3s)`, then find the relation between p, q, r and s.

Description : If `(2p+5q)/(2r+5s) = (4p-3q)/(4r-3s)`, then find the relation between p, q, r and s.

Last Answer : If `(2p+5q)/(2r+5s) = (4p-3q)/(4r-3s)`, then find the relation between p, q, r and s.

If P and ` Q^(2) ` are in direct proportion. If Q doubles then P becomes _______ .

Description : If P and ` Q^(2) ` are in direct proportion. If Q doubles then P becomes _______ .

Last Answer : If P and ` Q^(2) ` are in direct proportion. If Q doubles then P becomes _______ .

A person walks from point P to point Q along a straight road ¡n 10 minutes, then turns back and returns to point R which is midway

Description : A person walks from point P to point Q along a straight road ¡n 10 minutes, then turns back and returns to point R which is midway

Last Answer : A person walks from point P to point Q along a straight road n 10 minutes, then turns back and ... velocity of the person in going from P to R.

If P is moving away from a point and Q is moving towards a point then,

Description : If P is moving away from a point and Q is moving towards a point then,

Last Answer : If P is moving away from a point and Q is moving towards a point then, can their resultant be found using parallelogram law of vector addition?

The oxide of a metal (R ) can be reduced by the metal (P) and metal (R ) can reduce the oxide of metal (Q). Then the decreasing order of the reactivit

Description : The oxide of a metal (R ) can be reduced by the metal (P) and metal (R ) can reduce the oxide of metal (Q). Then the decreasing order of the reactivit

Last Answer : The oxide of a metal (R ) can be reduced by the metal (P) and metal (R ) can reduce the oxide of metal (Q). Then ... C. `R gt P gt Q` D. `Q gt P gt R`

For the chemical reaction P → Q, it is found that the rate of reaction doubles as the concentration of 'P' is doubled. If the reaction rate is proportional to Cp n , then what is the value of 'n' for this chemical reaction? (A) 0 (B) 1 (C) 2 (D) 3

Description : For the chemical reaction P → Q, it is found that the rate of reaction doubles as the concentration of 'P' is doubled. If the reaction rate is proportional to Cp n , then what is the value of 'n' for this chemical reaction? (A) 0 (B) 1 (C) 2 (D) 3

Last Answer : (B) 1

There is road besides a river. Two friends started from a place P, moved to a shopping mall situated at another place Q and then returned to P again. One of them moves on a cycle at a speed of 12 km/hr, while the other sails on a boat at a speed of 10 km/hr. If the river flows at the speed of 4 km/hr, which of the two friends will return to place P? A. Both B. Boater C. Cyclist D. None of these

Description : There is road besides a river. Two friends started from a place P, moved to a shopping mall situated at another place Q and then returned to P again. One of them moves on a cycle at a speed of 12 km/ ... which of the two friends will return to place P? A. Both B. Boater C. Cyclist D. None of these 

Last Answer : Answer-C Explanation: The cyclist moves both ways at a speed of 12khr so average speed fo the cyclist - 12 km/hr boat sailor moves downstream at 10+4 = 14km/hr and upstream 10- 4 = 6km/hr ... 8.4km/hr The average speed of cyclist is greater .so,cyclist comes first and return to place P.

P, Q and R can do a piece of work in 18,24 and 36 days on the second day and P on the third day, again R on the fourth day and so on. Then in how, many days will the work be completed? 1. 25 2. 24 3. 22 4. 20 5. 26

Description : P, Q and R can do a piece of work in 18,24 and 36 days on the second day and P on the third day, again R on the fourth day and so on. Then in how, many days will the work be completed? 1. 25 2. 24 3. 22 4. 20 5. 26

Last Answer : 2;

Five persons M, N, O, P and Q request a bank for safety lockers. There are only 6 lockers available which are numbered from 52 to 57. The conditions are as follows: I. M’s locker number is 2 less than O’s lockers number. II. Lockers of N and P are next to each other. If N’s locker number is 52 then M’s locker number could be a) 52 or 53 only b) 53 or 54 only c) 54 or 55 only d) 53 or 55 only

Description : Five persons M, N, O, P and Q request a bank for safety lockers. There are only 6 lockers available which are numbered from 52 to 57. The conditions are as follows: I. M's locker number is 2 less than O's lockers ... could be a) 52 or 53 only b) 53 or 54 only c) 54 or 55 only d) 53 or 55 only

Last Answer : c) 54 or 55 only

Five persons M, N, O, P and Q request a bank for safety lockers. There are only 6 lockers available which are numbered from 52 to 57. The conditions are as follows: I. M’s locker number is 2 less than O’s lockers number. II. Lockers of N and P are next to each other. 14. If Q’s locker number is 54 and N’s locker number is greater than Q’s locker number, then which locker number is not allocated? a) locker 52 b) locker 53 c) locker 54 d)locker 56

Description : Five persons M, N, O, P and Q request a bank for safety lockers. There are only 6 lockers available which are numbered from 52 to 57. The conditions are as follows: I. M's locker number is 2 less ... , then which locker number is not allocated? a) locker 52 b) locker 53 c) locker 54 d)locker 56

Last Answer : a) locker 52

P, Q, R, S and T are placed in the order of 1 to 5. S is at one of the extreme ends. Q and R are neighbors and T is 3rd to the right of P If Q is in the third position then who will be in fifth position? a) Both S and T b) Either P or Q c) Either S or T d) Both P and Q

Description : P, Q, R, S and T are placed in the order of 1 to 5. S is at one of the extreme ends. Q and R are neighbors and T is 3rd to the right of P If Q is in the third position then who will be in fifth position? a) Both S and T b) Either P or Q c) Either S or T d) Both P and Q

Last Answer : c) Either S or T