# Quantitative Aptitude Questions Daily Quiz Day – 33

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1) a) 17/13

b) 13/17

c) 13/7

d) 17/7

2) The sum of two numbers is 25 and sum of their square is 313. Calculate the numbers.

a) 15, 10

b) 18, 7

c) 11, 14

d) 12, 13

3) What is the value of a – b when a² + b² – 6a – 6b + 18 = 0 ?

a) 0

b) 3

c) 6

d) 9

4) The equation (1+a²) x² + 2abx + (b² – c²) = 0 has equal roots. What is the value of c² (1+a²) ?

a) a²

b) b²

c) c²

d) ab

5) A circle of diameter 26 cm having the two equal chords of length 10 cm is separated by a distance x cm. Calculate the distance x.

a) 20

b) 22

c) 24

d) 28

6) Two tangents PA and PB are drawn to a circle of centre O in such a way that AOB = 120°. Calculate the value of APB. a) 40°

b) 50°

c) 60°

d) 80°

7) The population of a town is 200000. It increases annually at the rate of 10%. What will be the population after 2 years?

a) 240000

b) 242000

c) 221000

d) 244000

8) The incomes of X and Y are in the ratio 5 : 6 and their expenditure are in the ratio 6 : 5. If X saves Rs 7000 and Y saves Rs 15000, then what will be the income (in Rs) of X?

a) 28000

b) 20000

c) 25000

d) 24000

9) X starts a business with Rs 80000. After 6 months Y joins X with Rs 100000. After 2 years, what will be the ratio of proﬁt of X and Y?

a) 16 : 15

b) 4 : 5

c) 8 : 9

d) 14 :  15

10) What is the average of all the one digit and two digit natural numbers?

a) 25

b) 40

c) 50

d) 99 Let the numbers, a & 25 – a

ATQ,

⇒ a² + (25–a)² = 313

⇒a² + 625 + a² – 50a = 313

⇒ 2a² – 50a + 312 = 0

⇒ a²– 25a + 156 = 0

⇒ a²– 13a – 12a+ 156 = 0

⇒a (a–13)–12 (a–13) = 0

⇒ (a–12)(a–13) = 0

a = 13, 12

Required number are, a = 12 or 13 & 25–a = 12

Numbers are 12 and 13.

a² + b² – 6a – 6b + 18 = 0

a² – 6a + 9 + b² – 6b + 9 = 0

(a–3)² + (b–3)² = 0

⇒ a = 3 & b = 3

⇒ a – b = 3 – 3 = 0

We have,

(1+a²) x² + 2abx + (b²–c²) = 0

for equal roots

⇒ B² – 4AC = 0

⇒ 4a²b² –4(1+a²) (b²–c²) = 0

⇒4a²b² – 4b² – 4a²b² + 4c² + 4a²c² = 0

⇒ 4c² (1+a²) = 4b²

⇒ c² (1+a²) = b² We have, PB = 5cm, OB = 13 cm

∴ OP = 12 cm

Similarly, OQ = 12cm

x = 12 + 12 = 24 cm

∠ PAO + ∠AOB + ∠OBP + ∠ APB = 360°

90° + 120° + 90° + ∠ APB = 360°

⇒∠APB = 60°   