Numerical Ability Test

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Numerical Ability Test - 9

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Weekly Quiz Competition

Question 1
5 / -1

The difference of two number is 2 and difference of their squares is 28. The sum of the number will be:

A
19

B
21

C
15

D
14

Solution

Given:
Difference of two numbers is 2
Difference of their squares is 28.
Concept used:
a2 - b2 = (a - b) (a + b)
Calculation:
Let the two number be x and y
⇒ x - y = 2 - - (i)
Difference of their squares
⇒ x2 - y2 = 28 - - (ii)
Using the formulae
⇒ 28 = 2 (x + y)
⇒ x + y = 28/2
⇒ x + y = 14
∴ The sun of the number is 14.
Question 2
5 / -1

If a + b + c = 10 and ab + bc + ca = 31, then the value of a³ +b³ + c³ - 3abc is

A
10

B
70

C
-10

D
-70

Solution

Given :
a + b + c = 10
ab + bc + ca = 31
Formula used :
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
a3 +b3 + c3 - 3abc = (a+ b + c) (a2 + b2 + c2 – ab – bc – ca)
Calculation :
We have,
ab + bc + ca = 31 ----(1)
a + b + c = 10 ----(2)
Squaring both side
(a + b + c) = 10²
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = (10)2
⇒ a2 + b2 + c2 + 2 × 31 = 100 [(From equation (1)]
⇒ a2 + b2 + c2 + 62 = 100
⇒ a2 + b2 + c2 = 38
By usnig the formula
a3 +b3 + c3 - 3abc = (a+ b + c) [a2 + b2 + c2 - (ab + bc + ca)]
⇒ a3 +b3 + c3 - 3abc = 10(38 - 31)
∴ a3 +b3 + c3 - 3abc = 70
Question 3
5 / -1

Simplify the following expression.
\(\frac{(375+125)^2-(125-375)^2}{375\times375-125\times125}\)

A
\(\frac{15}{8}\)

B
\(\frac{3}{4}\)

C
\(\frac{3}{2}\)

D
\(\frac{27}{28}\)

Solution

Given:
\(\frac{(375+125)^2-(125-375)^2}{375\times375-125\times125}\)
Concept Used:
(A + B)2 - (A - B)2 = 4AB
A2 - B2 = (A + B) (A - B)
Calculation:
Let 375 be A & 125 be B.
Now,
{(A + B)² - (B - A)²} ÷ (A × A - B × B)
⇒ {(A + B)2 - (A - B)2} ÷ (A² - B²)
⇒ (4 × AB) ÷ {(A + B) (A - B)}
⇒ (4 × 375 × 125) ÷ {(375 + 125) (375 - 125)}
⇒ \(\frac{4 × 375 × 125 } {500 \times 250}\)
⇒ 375/250
⇒ 3/2
∴ The value of given expression in 3/2.
Question 4
5 / -1

if x + y = 3 and \(\frac{1}{x}+\frac{1}{y}=-\frac{3}{10}\), then the value of (x² + y²) is:

A
28

B
34

C
29

D
26

Solution

Given:
x + y = 3
1/x + 1/y = - (3/10)
Concept Used:
(A + B)²= A²+ 2AB + B²
Calculation:
1/x + 1/y = - (3/10)
⇒ (x + y)/xy = - (3/10)
⇒ xy = (-10)
Now,
x + y = 3
⇒ (x + y)² = 9 (Squaring both side)
⇒ x2 + y2 + 2xy = 9
⇒ x2 + y2 = 9 - 2xy
⇒ x2 + y2 = 9 – 2 × (-10)
⇒ x² + y²= 29
∴ The required value of x² + y² is 29.
Question 5
5 / -1

If a + b = 8 and a² + b² = 34 (a > b), then a² - b² will be

A
2

B
4√2

C
16

D
None of these
Solution
Given
a + b = 8 ----(1)
a2 + b2 = 34 ----(2)
Formula used:
- (a + b)2 = a2 + 2ab + b2
- (a - b)2 = a2 - 2ab + b2
- a2 - b2 = (a + b)(a - b)
Calculation:
∵ (a + b)2 = a2 + 2ab + b2
⇒ 8²= 34 + 2ab
⇒ 2ab = 30 ----(3)
⇒ ab = 15
∵ (a - b)2 = a2 - 2ab + b2
⇒ (a - b)2 = 34 - 2 × 15 = 4
⇒ a - b = 2 ----(4)
∵ a2 - b2 = (a + b)(a - b)
Hence, required value suing the equation (1) and (4)
a2 - b2 = 8 × 2
⇒ a2 - b2 = 16
Question 6
5 / -1

Find the value of 101³.

A
1003031

B
1030301

C
1303001

D
1130301

Solution

Given:
The given number is 101³.
Formula used:
(a + b)³ = a³ + b³ + 3ab(a + b)
Calculation:
(101)³
⇒ (100 + 1)³
⇒ (100)³ + 1³ + [3 × 100 × 101]
⇒ 1000000 + 1 + 30300
⇒ 1030301
Question 7
5 / -1

Select the most appropriate option to solve the equation: 6719² - 3281² = ?

A
27440000

B
26620000

C
34380000

D
25540000

Solution

Given:
(6719)² - (3281)²
Formula used:
a² - b² = (a - b) (a + b)
Calculation:
(6719)2 - (3281)²
⇒ (6719 - 3281) (6719 + 3281)
⇒ 3438 × 10000
⇒ 34380000
∴ The value of (6719)² - (3281)² is 34380000.
Question 8
5 / -1

If (a - b) = 3, ab = 18, find the value of a³ - b³

A
178

B
167

C
189

D
198

Solution

Given
(a - b) = 3 and ab = 18
Formula used
(a - b)³ = a³ - b³- 3ab(a - b)
Calculation
(3)³ = a³- b³- 3 × 18 × 3
⇒ 27 = a³- b³ - 162
⇒ 27 + 162 = a³ - b³
⇒ 189 = a³ - b³
∴ The value of a³- b³ is 189
Question 9
5 / -1

If a + b = 9 and ab = 20, then find the value of (a³ + b³).

A
150

B
210

C
189

D
190

Solution

Given:
A + b = 9 and ab = 20
Formula:
a³ + b³ = (a + b)³ – 3ab(a + b)
Calculation:
a³ + b³ = (a + b)³ – 3ab(a + b)
⇒ a³ + b³ = 9³ – 3 × 20 × 9
⇒ a³ + b³ = 729 - 540
⇒ a³ + b³ = 189
Smart Trick
Take a = 5 and b = 4 (a + b = 5 + 4 = 9 and ab = 5 × 4 = 20)
∴ a³ + b³ = 5³ + 4³ = 125 + 64 = 189.
Question 10
5 / -1

The factors of x² + 8x + 12 are:
A. (x + 6) (x + 2)
B. (x - 6) (x + 2)
C. (x - 4)²
D. (x + 6)(x - 2)

A
B

B
A

C
C

D
D

Solution

Calculation:
x² + 8x + 12
⇒ x² + 6x + 2x + 12
⇒ x (x + 6) + 2 (x + 6)
⇒ (x + 6) (x + 2)

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Re-Attempt Weekly Quiz Competition

Numerical Ability Test - 9

Result

Numerical Ability Test - 9

Score

-

Rank

-

SHARING IS CARING

Question 1

19

21

15

14

Solution

Question 2

10

70

-10

-70

Solution

Question 3

\(\frac{15}{8}\)

\(\frac{3}{4}\)

\(\frac{3}{2}\)

\(\frac{27}{28}\)

Solution

Question 4

28

34

29

26

Solution

Question 5

2

4√2

16

None of these

Solution

Question 6

1003031

1030301

1303001

1130301

Solution

Question 7

27440000

26620000

34380000

25540000

Solution

Question 8

178

167

189

198

Solution

Question 9

150

210

189

190

Solution

Question 10

B

A

C

D

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

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