**Dear Aspirants,** **Our SSC Crackers team** is providing a new series of Quantitative Aptitude Questions for **Upcoming Exam **so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

**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 3 ^{13} by 5.**

a) 0

b) 1

c) 2

d) 3

**6) If a = (11 + 2√18) ^{½} and b = (11 – 2√18)^{½}, find a^{3} + b^{3}.**

a) 86

b) 90

c) 110

d) 135

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

**(m ^{2}−n^{2}−p^{2}−2np)/(m^{2}+n^{2}−2mn−p^{2})**

a) 1

b) 1.5

c) 2

d) 2.5

**8) If x = 3 + √8, find the value of x ^{2} + 1/ x^{2}**

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

**Answers :**

**1) Answer: C) **

Given, a – b = 1.5

Squaring both sides, we get,

a^{2 }+ b^{2 }– 2ab = 1.5^{2}

a^{2 }+ b^{2 }= 2.25 + 2 × 17.5

a^{2 }+ b^{2 }= 37.25

a^{2 }+ b^{2 }– 2ab + 2ab = 37.25

a^{2 }+ b^{2 }+ 2ab = 37.25 + 2 × 17.5

(a– b)^{2 }= 72.25

(a – b) = √72.25

(a – b) = 8.5

**2) Answer: A) **

(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

**3) Answer: C) **

(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

**4) Answer: A) **

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

**5) Answer: D) **

3/5 = (-2)

3^{2}/5 = (-1)

(3^{2})^{6}/5 = (-1)^{6 }= 1

(3^{13})/5 = [(3^{2})^{6}.3]/5 = -2 = (-2) + 5 = 3

**6) Answer: B) **

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

a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b)

=> a^{3} + b^{3} = 6^{3} – 3 * 7 * 6 = 216 – 126

=> a^{3} + b^{3} = 90

**7) Answer: B) **

(m^{2}−n^{2}−p^{2}−2np)/(m^{2}+n^{2}−2mn−p^{2})

=[m^{2}−(n^{2}+p^{2}+2np)]/[(m^{2}+n^{2}−2mn)−p^{2}]

=[m^{2}−(n+p)^{2}]/[(m−n)^{2}−p^{2}]

=[(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

**8) Answer: C) **

x=3+√8

x^{2} = (3+√8)^{2} = 9 + 8 + 6√8 = 17 + 6√8

1/x^{2} = 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

**9) Answer: A)**

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

**10) Answer: C) **

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