Explore topic-wise InterviewSolutions in Current Affairs.

Solution

117% of 459.88 - 162% of 143.02.
~117% of 460-162% of 143.
=$\FRAC{117\times460}{100}-\frac{162\times143}{100}$.
=$\frac{53820}{100}-\frac{23166}{100}$.
=$538.20-231.66$.
~$306$.
Hence, Option D is CORRECT.

Solution

According to Vijay’s mother, Vijay reached Mumbai after 13th January, 2013 but before 19th January, 2013

=> Date at which Vijay reached Mumbai (JAN, 2013) = {14,15,16,17,18}

According to Vijay, he reached Mumbai after 16th January, 2013 but before 22ND January, 2013

=> Date at which Vijay reached Mumbai (Jan, 2013) = {17,18,19,20,21}

Since both of them correct, thus correctdate on which Vijay reached Mumbai is the INTERSECTION of both SETS

= 17 or 18 January

Still, we cannot conclude to a specific date, thus the correct data cannot be determined.

=> Ans - (E)

Solution

Let UNKNOWN QUANTITY be 'x'.
3567.43 + 2788.17 - x = 1379.56.
6355.60-x=1379.56.
x=6355.60-1379.56.
x=4976.04.
Hence, OPTION B is CORRECT.

it from PREVIOUS year ssc papers, OPTION 1 is the right ANSWER

I THINK OPTION 1 is the RIGHT ANSWER

Solution

12 men can finish the project in 20 days.

=> 1 day work of 1 man = $\frac{1}{12 \times 20} = \frac{1}{240}$

Similarly, => 1 day work of 1 woman = $\frac{1}{18 \times 16} = \frac{1}{288}$

=> 1 day work of 1 CHILDREN = $\frac{1}{24 \times 18} = \frac{1}{432}$

8 WOMEN and 16 children worked for 9 days

=> Work done in 9 days = $9 \times (8 \times \frac{1}{288}) + (16 \times \frac{1}{432})$

= $9 \times (\frac{1}{36} + \frac{1}{27}) = 9 \times \frac{7}{108}$

= $\frac{7}{12}$

=> Work left = $1 - \frac{7}{12} = \frac{5}{12}$

$\therefore$ Number of days taken by 10 men to complete the remaining work

= $\frac{\frac{10}{240}}{\frac{5}{12}} = \frac{1}{24} \times \frac{12}{5} = \frac{1}{10}$

Thus, 10 men will complete the remaining the work in 10 days.

Solution

Initially, number of boys = 80 and number of girls = 75

Average weight of boys = 64 kg and average weight of girls = 70 kg

Now, 60% of the girls and 30% of the boys LEAVE

=> Boys LEFT = $\frac{100 - 30}{100} \TIMES 80 = 56$

Girls left = $\frac{100 - 60}{100} \times 75 = 30$

Since, average weight of the boys and the girls remains constant throughout

$\therefore$ New average weight of the class

= $\frac{(56 \times 64) + (30 \times 70)}{56 + 30} = \frac{3584 + 2100}{86}$

= $\frac{5684}{86} = 66.09$ kg

Solution

73% of 5800 = (70+3)% of 5800 = 4060+174 = 4234

69% of 240 = (70-1)% of 240 =168-2.4 = 165.6

Hence, ANSWER will be = 4234 - 165.6 = 4068.4

Solution

25% of 960 + 55% of 740.
=$\FRAC{25\times960}{100}+\frac{55\times740}{100}$.
=$240+407$.
=$647$.
HENCE, Option B is CORRECT.

Solution

Expression : 432.62 - 269.21 $\div$(11.9 % of 78) =?

= $432 - (270 \times \frac{1}{\frac{12}{100} \times 78})$

= $432 - (270 \times \frac{100}{936})$

= $432 - 29 = 403$

$\approx 400$

Solution

Expression : ? % of (767 ÷ 6) = 7.889²

=> $x \%$ of $\frac{768}{6} = (8)^2$

=> $\frac{x}{100} \times 128 = 64$

=> $x = 64 \times \frac{100}{128}$

=> $x = \frac{100}{2} = 50$

Solution

Let EVEN numbers be '2n' and '2n+2'.
According to question,
$2n\times(2n+2)=7568$.
$4n(n+1)=7568$.
$n(n+1)=1892$.
$n=43$.
HENCE, numbers are 86 and 88 respectively.
Sum of numbers=86+88.
=174.
150% of sum of numbers=$1.5\times174$.
$=261$.
Hence, Option B is correct.

Solution

Initially, number of boys = 40 and number of girls = 50

Average weight of boys = 65 kg and average weight of girls = 60 kg

Now, 40% of the girls and 50% of the boys leave

=> Boys left = $\FRAC{100 - 50}{100} \times 40 = 20$

Girls left = $\frac{100 - 40}{100} \times 50 = 30$

SINCE, average weight of the boys and the girls remains constant throughout

$\THEREFORE$ New average weight of the class

= $\frac{(20 \times 65) + (30 \times 60)}{20 + 30} = \frac{1300 + 1800}{50}$

= $\frac{3100}{50} = 62$ kg

Solution

Let number be 'N'.
It is given,
4/5th of n-20% of n=2499.
$\frac{4}{5}n-\frac{1}{5}n=2499$.
$\frac{3}{5}n=2499$.
$n=4165$.
2/7th of n=$\frac{2}{7}4165$.
=1190.
Hence, OPTION B is CORRECT.

Solution

Jar A has 78 litres of mixture of milk and water in the respective RATIO of 6 : 7

=> Quantity of milk in Jar A = $\frac{6}{13} \times 78 = 36$ litres

Quantity of water in Jar A = $78 - 36 = 42$ litres

26 litres of the mixture was taken out from Jar A, i.e., $\frac{26}{78} = (\frac{1}{3})^{rd}$

=> Milk left = $36 - \frac{1}{3} \times 36 = 24$

Water left = $42 - \frac{1}{3} \times 42 = 28$

Let milk added to jar A = $x$ litres

Acc. to QUES, => $\frac{24 + x}{28} = \frac{60}{40}$

=> $\frac{24 + x}{28} = \frac{3}{2}$

=> $48 + 2X = 84$

=> $2x = 84 - 48 = 36$

=> $x = \frac{36}{2} = 18$ litres

Solution

Let the five consecutive odd numbers in increasing order = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$

SUM of these numbers = $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 225$

=> $5x = 225$

=> $x = \frac{225}{5} = 45$

Thus, the odd numbers are = 41 , 43 , 45 , 47 , 49

Let another series of even numbers in increasing order = $(y-4) , (y-2) , (y) , (y+2) , (y+4)$

Also, $43 = (y - 2) - 15$

=> $y = 43 + 15 + 2 = 60$

Thus, HIGHEST number of the even series = 60 + 4 = 64

$\therefore$ 60% of 64 = $\frac{60}{100} \TIMES 64 = 38.4$

Solution

Let speed of the train = $20x$ m/s

Now, 75% of the speed = $\frac{75}{100} \times 20x = 15x$ m/s

LENGTH of train = 150 m

Time taken to CROSS the pole = 12 sec

Using, $speed = \frac{distance}{time}$

=> $15x = \frac{150}{12}$

=> $x = \frac{10}{12} = \frac{5}{6}$

Length of platform = $l$ m

Acc. to ques, => $20x = \frac{150 + l}{24}$

=> $20 \times \frac{5}{6} = \frac{150 + l}{24}$

=> $\frac{50}{3} = \frac{150 + l}{24}$

=> $150 + l = 400$

=> $l = 400 - 150 = 250$ m

Solution

Expression : 459.85 + 519.82 = ?% of 1399.92

=> $460 + 520 = \frac{X}{100} \TIMES 1400$

=> $980 = 14x$

=> $x = \frac{980}{14} = 70$

Solution

Expression : 83% of 6242 X 12% of 225 = ?

= $(\FRAC{83}{100} \TIMES 6242) \times (\frac{12}{100} \times 225)$

= $5180.86 \times 27$

= $139883.22$

Monetary policy is the CORRECT answer as PER the SSC answer KEY

Solution

273 $\DIV$ 3 - 9.1.
=91-9.1.
=81.9.
Hence, OPTION AIS CORRECT.

i THINK OPTION 4 is CORRECT

Its TOUGH but OPTION 2 SEEMS CORRECT

He had forgotten the magic SPELL and all Govind’s hard work was in VAIN SEEMS CORRECT.

is FREQUENT ABSENT is the CORRECT ANSWER?? am i RIGHT?

This question was ASKED some where in PREVIOUS year PAPERS of ssc, and CORRECT ANSWER was option 3

Most of the PEOPLE which SEEMS CORRECT.

Its very SIMPLE QUESTION STANDS is the CORRECT ANSWER.

it from PREVIOUS year ssc papers, SUBSTANTIALLY is the RIGHT ANSWER

Its very SIMPLE QUESTION 2 is the CORRECT ANSWER.