Quantitative Aptitude Questions Daily Quiz Day – 28

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1) If (a – b) = 1.5, and ab = 17.5, then find (a + b).

a) 7.5

b) 8

c) 8.5

d) 9

2) If (x – 5)2 + (y + 2)2 + (z – 4)2 = 0, find the value of (2x – 2y + 3z).

a) 26

b) 24

c) 28

d) 22

3) If (a + b + c + d) = 16, then find the maximum value of (a + 2) (b – 1) (c – 2) (d + 5).

a) 81

b) 256

c) 625

d) 1296

4) Let x denote the greatest 4-digit number which when divided by 6, 7, 8, 9 and 10 leaves a remainder of 4, 5, 6, 7 and 8 respectively. Then, the sum of the four digits of x is:

a) 25

b) 18

c) 20

d) 22

5) Find the remainder on dividing 313 by 5.

a) 0

b) 1

c) 2

d) 3

6) If a = (11 + 2√18)½ and b = (11 – 2√18)½, find a3 + b3.

a) 86

b) 90

c) 110

d) 135

7) If m = p = 0.5, n = 0.2, then the value of

(m2−n2−p2−2np)/(m2+n2−2mn−p2)

a) 1

b) 1.5

c) 2

d) 2.5

8) If x = 3 + √8, find the value of x2 + 1/ x2

a) 38

b) 36

c) 34

d) 30

9) If the graph of the equations 3x + 2y = 18 and 3y – 2x = 1 intersect at the point (p,q), then the value of p + q is

a) 7

b) 6

c) 5

d) 4

10) The equations 3x + 4y = 10, – x + 2y = 0 have the solution (a,b), the value of a + b is

a) 1

b) 2

c) 3

d) 4

Given, a – b = 1.5

Squaring both sides, we get,

a2 + b2 – 2ab = 1.52

a2 + b2 = 2.25 + 2 × 17.5

a2 + b2 = 37.25

a2 + b2 – 2ab + 2ab = 37.25

a2 + b2 + 2ab = 37.25 + 2 × 17.5

(a– b)2 = 72.25

(a – b) = √72.25

(a – b) = 8.5

(x – 5)2 + (y + 2)2 + (z – 4)2 = 0

=> x – 5 = 0, y + 2 = 0, z – 4 = 0 [If sum of two or more positive numbers is zero that means the numbers themselves are zero]

=> x = 5, y = -2, z = 4

Now,

(2x – 2y + 3z) = 2 x 5 – 2 x (-2) + 3 x 4

= 10 + 4 + 12

= 26

(a + 2) + (b – 1) + (c – 2) + (d + 5)= (a + b + c + d) + 4 = 16 + 4 = 20

For a given sum, product is maximum when all the terms are equal, so,

(a + 2) = (b – 1) = (c – 2) = (d + 5) = 20/4 = 5

Therefore, Max [(a + 2)(b – 1)(c – 2)(d + 5)] = 5 × 5 × 5 × 5 = 625

LCM of (6, 7, 8, 9, 10) = 2520

The largest 4-digit number divisible by 6, 7, 8, 9 and 10 = 2520 X 3 = 7560

Hence, when 7558 is divided by 6, 7, 8, 9 and 10 leaves a remainder of 4, 5, 6, 7 and 8 respectively

Sum of the digits = 25

3/5 = (-2)

32/5 = (-1)

(32)6/5 = (-1)6 = 1

(313)/5 = [(32)6.3]/5 = -2 = (-2) + 5 = 3

11 + 2√18 = 9 + 2 + 2 x 3 √2 = (3 + √2)2

11 – 2√18 = 9 + 2 – 2 x 3 √2 = (3 – √2)2

So, a = 3 + √2 and b = 3-√2

∴ a + b = (3 + √2) + (3 – √2) = 6

a * b = (3 + √2) * (3 – √2) = 9 – 2 = 7

a3 + b3 = (a + b)3 – 3ab(a + b)

=> a3 + b3 = 63 – 3 * 7 * 6 = 216 – 126

=> a3 + b3 = 90

(m2−n2−p2−2np)/(m2+n2−2mn−p2)

=[m2−(n2+p2+2np)]/[(m2+n2−2mn)−p2]

=[m2−(n+p)2]/[(m−n)2−p2]

=[(m+n+p)(m−n−p)]/[(m−n+p)(m−n−p)]

=(m+n+p)/(m−n+p)

= (0.5+0.2+0.5)/(0.5 -0.2+0.5)

= 1.2/0.8 = 1.5

x=3+√8

x2 = (3+√8)2 = 9 + 8 + 6√8 = 17 + 6√8

1/x2 = 1/(17 + 6√8) = (17 – 6√8) / ( (17 + 6√8) – (17 + 6√8))

= (17 – 6√8) / (289 – 288) = (17 – 6√8)

Required value = 17 + 6√8 + 17 – 6√8 = 34

3x + 2y = 18 ……(i)

3y – 2x = 1 …….(ii)

By equation (i)×2+(ii)×3 gives

13y = 39

⇒ y = 3

Putting y = 3 in (ii)

3(3) – 2x = 1 ⇒ x = 4

(p, q) = (4, 3)

And hence, p + q = 7

3x + 4y = 10 …..(i)

– x + 2y = 0

⇒ x = 2y

From equation (i),  3 × 2y + 4y = 10

⇒ 10y = 10

y=1;

So x = 2;

(a,b) = (2,1); a+b = 2 + 1 = 3