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Q:

${(31)}^{31} \times {(31)}^{- 27} = ?$

A) 961 x 861	B) 961 x 961
C) 961 x 2	D) 29791

Answer & Explanation Answer: B) 961 x 961

Explanation:

We know that $a^{m} \times a^{n} = a^{m + n}$

$∴ {(31)}^{31 - 27}$ = $31^{4}$ = 961 x 961.

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Filed Under: Numbers
Exam Prep: CAT
Job Role: Bank PO

Q:

A,B and C enter into a partnership in the ratio $\frac{7}{2} : \frac{4}{3} : \frac{6}{5}$ . After 4 months, A increases his share by 50%. If the total profit at the end of one year be Rs. 21600, then B's share in the profit is :

A) Rs. 4000	B) Rs. 6000
C) Rs. 9000	D) Rs. 3000

Answer & Explanation Answer: A) Rs. 4000

Explanation:

Given ratio of initial investments = $\frac{7}{2} : \frac{4}{3} : \frac{6}{5}$ = 105 : 40 : 36.

Let the initial investments be 105x, 40x and 36x.

$∴ A : B : C = (105 x \times 4 + \frac{150}{100} \times 105 x \times 8) : (40 x \times 12) : (36 x \times 12)$

= 1680x : 480x : 432x = 35 : 10 : 9.

Hence, B's share = $(21600 \times \frac{10}{54})$ = Rs. 4000.

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Filed Under: Partnership
Exam Prep: CAT
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Q:

If a+b=5 and 3a+2b=20,then (3a+b)will be :

A) 20	B) 15
C) 25	D) 30

Answer & Explanation Answer: C) 25

Explanation:

a+b=5 ...(1) and 3a+2b=20 ...(2)

Multiplying (1) by 2 and subtracting from (2), we get : a=10.

Putting a=10 in (1), we get : b=-5

Therefore, (3a+b) = 3 x 10+(-5)=30-5=25.

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Filed Under: Simplification
Exam Prep: CAT
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Q:

A sales boy is allowed $4 \frac{1}{2}$ % discount on the total sales made by him plus a bonus of $\frac{1}{4}$ % on the sales over Rs.12,000.If his total earnings were Rs.2400, then his total sales(in Rs.) are ?

A) Rs.60,145.23	B) Rs.54,285.54
C) Rs.52,147.85	D) Rs.50,589.47

Answer & Explanation Answer: D) Rs.50,589.47

Explanation:

Let the total earnings be Rs.x

Then $4 \frac{1}{2}$ % of x + $\frac{1}{4}$ % of (x-12000) =2400

$\Rightarrow \frac{9}{2} \times \frac{x}{100} + \frac{1}{4} \times \frac{x - 12000}{100} = 2400$

=> x=50,589.47

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Filed Under: Percentage
Exam Prep: CAT
Job Role: Bank Clerk

Q:

Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?

(1) $K_{2} O < N a_{2} O < A l_{2} O_{3} < M g O$

(2) $A l_{2} O_{3} < M g O < N a_{2} O < K_{2} O$

(3) $M g O < K_{2} O < A l_{2} O_{3} < N a_{2} O$

(4) $N a_{2} O < K_{2} O < M g O < A l_{2} O_{3}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: B) Option 2

Explanation:

While moving from left to right in periodic table basic character of oxide of elements will decrease.

$∴ \to_{I n c r e a \sin g b a s i c s t r e n g t h}^{A l_{2} O_{3} < M g O < N a_{2} O}$

and while descending in the group basic character of corresponding oxides increases.

$∴ \to_{I n c r e a \sin g b a s i c s t r e n g t h}^{N a_{2} O < K_{2} O}$

$∴ C o r r e c t O r d e r i s A l_{2} O_{3} < M g O < N a_{2} O < K_{2} O$

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Filed Under: Chemistry
Exam Prep: AIEEE

Q:

Boron cannot form which one of the following anions?

(1) $B O_{2}^{-}$

(2) $B F_{6}^{3 -}$

(3) $B H_{4}^{-}$

(4) $B (O H)_{4}^{-}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: B) Option 2

Explanation:

Due to absence of low lying vacant d orbital in B, $s p^{3} d^{2}$ hybridization is not possible hence $B F_{6}^{3 -}$ will not formed.

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Filed Under: Chemistry
Exam Prep: AIEEE

Q:

Among the following the maximum covalent character is shown by the compound

(1) $M g C l_{2}$

(2) $F e C l_{2}$

(3) $S n C l_{2}$

(4) $A l C l_{3}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: D) Option 4

Explanation:

According to Fajans rule, cation with greater charge and smaller size favours covalency.

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Filed Under: Chemistry
Exam Prep: AIEEE

Q:

The reduction potential of hydrogen half-cell will be negative if :
(1) $P (H_{2})$ = 2 atm and [H+] = 2.0 M
(2) $P (H_{2})$ = 1 atm and [H+] = 2.0 M
(3) $P (H_{2})$ = 1 atm and [H+] = 1.0 M
(4) $P (H_{2})$ = 2 atm and [H+] = 1.0 M

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: D) Option 4

Explanation:

$H^{+} - e^{-} \to \frac{1}{2} H_{2}$

Apply Nernst equation

$E = 0 - \frac{0.059}{1} \log (\frac{P H_{2}^{\frac{1}{2}}}{[H^{+}]})$

$E = - \frac{0.059}{1} \log (\frac{2^{\frac{1}{2}}}{1})$

Therefore E is Negative

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Filed Under: Chemistry
Exam Prep: AIEEE

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