Q1. The Mean Weight Of A Class Of 35 Students Is 45 Kg. If The Weight Of The Teacher Be Included, The Mean Weight Increases By 500 G. Find The Weight Of The Teacher?

Mean weight of 35 students = 45 kg.

Total weight of 35 students = (45 × 35) kg = 1575 kg.

Mean weight of 35 students and the teacher (45 + 0.5) kg = 45.5 kg.

Total weight of 35 students and the teacher = (45.5 × 36) kg = 1638 kg.

Weight of the teacher = (1638 - 1575) kg = 63 kg.

Hence, the weight of the teacher is 63 kg.

Q2. In A Fort, There Are 1200 Soldiers. If Each Soldier Consumes 3 Kg Per Day, The Provisions Available In The Fort Will Last For 30 Days. If Some More Soldiers Join, The Provisions Available Will Last Fo

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)

= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available

=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = [(1200)(30)(3)] / (25)(2.5) – 1200 => x = 528.

Q3. Two Men Amar And Bhuvan Have The Ratio Of Their Monthly Incomes As 6 :

Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.

Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.

Savings of Bhuvan every month = 1/4(5x)

= (His income) – (His expenditure) = 5x – 2y.

=> 5x = 20x – 8y => y = 15x/8.

Ratio of savings of Amar and Bhuvan

= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4

= 3 : 10.

Q4. 12l : 24x :: 5e : __?

12L : 24X :: 5E : __

L is 12th letter and 12 * 2 = 24

The 24th letter is X.

Similarly, E is the 5th letter and 5 * 2 = 10

The 10th letter is J.

Q5. The Average Height Of 30 Boys Was Calculated To Be 150 Cm. It Was Detected Later That One Value Of 165 Cm Was Wrongly Copied As 135 Cm For The Computation Of The Mean. Find The Correct Mean?

Calculated average height of 30 boys = 150 cm.

Incorrect sum of the heights of 30 boys

= (150 × 30)cm

= 4500 cm.

Correct sum of the heights of 30 boys

= (incorrect sum) - (wrongly copied item) + (actual item)

= (4500 - 135 + 165) cm

= 4530 cm.

Correct mean = correct sum/number of boys

= (4530/30) cm

= 151 cm.

Hence, the correct mean height is 151 cm.

Q6. What Is The Least Number To Be Subtracted From 11, 15, 21 And 30 Each So That Resultant Numbers Become Proportional?

Let the least number to be subtracted be x, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.

<=> (11 – x) : (15 – x) = (21 – x) : (30 – x)

=> (11 – x)(30 – x) = (15 – x)(21 – x)

From the options, when x = 3

=> 8 * 27 = 12 * 18

Q7. A Man, A Woman And A Boy Can Complete A Job In 3, 4 And 12 Days Respectively. How Many Boys Must Assist 1 Man And 1 Woman To Complete The Job In 1/4 Of A Day?

(1 man + 1 woman)’s 1 day work = (1/3 + 1/4) = 7/12 Work done by 1 man and 1 woman in 1/4 day = (7/12 * 1/4) = 7/48

Remaining work = (1 – 7/48) = 41/48

Work done by 1 boy in 1/4 day = ( 1/12 * 1/4) = 1/48

Number of boys required = 41/48 * 41 = 41

Q8. How Many Values Of C In Equation X^2-5x+c Result In Rational Roots Which Are Integers?

Mean weight of 7 boys = 56 kg.

Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg = 338 kg.

Weight of the 7th boy = (total weight of 7 boys) – (total weight of 6 boys) = (392 – 338) kg = 54 kg.

Q9. A Cricketer Has A Mean Score Of 58 Runs In Nine Innings. Find Out How Many Runs Are To Be Scored By Him In The Tenth Innings To Raise The Mean Score To 61?

Mean score of 9 innings = 58 runs.

Total score of 9 innings = (58 x 9) runs = 522 runs.

Required mean score of 10 innings = 61 runs.

Required total score of 10 innings = (61 x 10) runs = 610 runs.

Number of runs to be scored in the 10th innings

= (total score of 10 innings) - (total score of 9 innings)

= (610 -522) = 88.

Hence, the number of runs to be scored in the 10th innings = 88.

Q10. Ax^2+bx+c = 0 Has Two Roots X1 And X

mod x1 = mod x2 => x1 & x2 have same value but opposite in sign.

in this case b=0 eqn is ax^2+c = 0 => x=sqrt(-c/a) for x to be real (-c/a) should +ve => c & a are of opposite sign so c>a or a>c for ex x^2-4=0 => x1=-2,

x2=2 & modx1=modx2=2 [a>c] or may -x^2+4=0 => x1=-2,x2=2 & modx1=modx2=2 (d) none

Q11. How Many Values Of C In The Equation X^3-5x+c Result In Rational Roots Which Are Integers?

x^3-5x+c=0 if x=1,

1-5+c=0 or c=4 if x=-1,

-1+5+c=0 or c=-4 if x=2, 8-10+c=0 or

c=2 if x=-2, -8+10+c=0 or c=-2 we see that,

for c=-2,2,-4,4

we get x=-2,2,-1,1 etc i.e we get integral roots of x for infinite values of c. d)infinite

Q12. The Mean Of 14 Numbers Is

Let the given numbers be x1, x2, x3, ….. x14.

Then, the mean of these numbers = x1 + x2 + x3+ ….. x14/14

Therefore, (x1 + x2 + x3 + ….. x14)/14 = 6

⇒ (x1 + x2 + x3 + ….. x14) = 84 ………………. (A)

The new numbers are (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)

Mean of the new numbers

= (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)/14

= (x1 + x2 + x3 + ….. x14) + 42

= (84 + 42)/14, [Using (A)]

= 126/14

= 9

Hence, the new mean is 9.

Q13. Find Remainder Of (9^1+9^2+.........+9^n)/6 N Is Multiple Of 11?

9/6 remainder is 3 9^2/6 remainder is 3 9^3/6 remainder is 3 9 to the power of any number when divided by 6 ,

the remainder will always be 3.

Now, ( 3+3+3 ……11 times)/6 =(3*11)/6;

therefore the remainder will be 3.

If we take the even multiple of 11, then remainder will be zero.

Therefore answer is cannot be determine

Q14. The Mean Of Five Numbers Is 2

Mean of 5 numbers = 28.

Sum of these 5 numbers = (28 x 5) = 140.

Mean of the remaining 4 numbers = (28 - 2) =26.

Sum of these remaining 4 numbers = (26 × 4) = 104.

Excluded number

= (sum of the given 5 numbers) - (sum of the remaining 4 numbers)

= (140 - 104)

= 36.

Hence, the excluded number is 36.

Q15. The Mean Of Eight Numbers Is 2

Let the given numbers be x1, x2, . . ., x8.

Then, the mean of these numbers = (x1 + x2 + ...+ x8)/8.

Therefore, (x1 + x2+...+x8)/8 = 25

⇒ (x1 + x2 + ... + x8) = 200 ……. (A)

The new numbers are (x1 - 5), (x2 - 5), …… ,(x8 - 5)

Mean of the new numbers = {(x1 - 5) + (x1 - 5) + …… + (x8 - 5)}/8

= [(x1 + x2 + ... + x8) - 40]/8

= (200 - 40)/8, [using (A)]

= 160/8

= 20

Hence, the new mean is 20.

Q16. The Mean Of 25 Observations Is 3

Mean of the first 13 observations = 32.

Sum of the first 13 observations = (32 × 13) = 416.

Mean of the last 13 observations = 39.

Sum of the last 13 observations = (39 × 13) = 507.

Mean of 25 observations = 36.

Sum of all the 25 observations = (36 × 25) = 900.

Therefore, the 13th observation = (416 + 507 - 900) = 23.

Hence, the 13th observation is 23.

Q17. The Mean Of 16 Items Was Found To Be 3

Calculated mean of 16 items = 30.

Incorrect sum of these 16 items = (30 × 16) = 480.

Correct sum of these 16 items

= (incorrect sum) - (sum of incorrect items) + (sum of actual items)

= [480 - (22 + 18) + (32 + 28)]

= 500.

Therefore, correct mean = 500/16 = 31.25.

Hence, the correct mean is 31.25.

Q18. The Mean Weight Of A Group Of Seven Boys Is 56 Kg. The Individual Weights (in Kg) Of Six Of Them Are 52, 57, 55, 60, 59 And 5

Mean weight of 7 boys = 56 kg.

Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg

= 338 kg.

Weight of the 7th boy = (total weight of 7 boys) - (total weight of 6 boys)

= (392 - 338) kg

= 54 kg.

Hence, the weight of the seventh boy is 54 kg.

Q19. The Weights Of Three Boys Are In The Ratio 4 : 5 :

Let the weights of the three boys be 4k, 5k and 6k respectively.

4k + 6k = 5k + 45

=> 5k = 45 => k = 9

Therefore the weight of the lightest boy

= 4k = 4(9) = 36 kg.

Q20. The Aggregate Monthly Expenditure Of A Family Was $ 6240 During The First 3 Months, $ 6780 During The Next 4 Months And $ 7236 During The Last 5 Months Of A Year. If The Total Saving During The Year I

Total expenditure during the year

= $[6240 × 3 + 6780 × 4 + 7236 × 5]

= $ [18720 + 27120 + 36180]

= $ 82020.

Total income during the year = $ (82020 + 7080) = $ 89100.

Average monthly income = (89100/12) = $7425.

Hence, the average monthly income of the family is $ 7425.

Q21. A 270 M Long Train Running At The Speed Of 120 Km/hr Crosses Another Train Running In Opposite Direction At The Speed Of 80 Km/hr In 9 Sec. What Is The Length Of The Other Train?

Relative speed = 120 + 80 = 200 km/hr.

= 200 * 5/18 = 500/9 m/sec.

Let the length of the other train be x m.

Then, (x + 270)/9 = 500/9 => x = 230.

Q22. In 100 M Race, A Covers The Distance In 36 Seconds And B In 45 Seconds. In This Race A Beats B By?

In 100 m race, A covers the distance in 36 seconds and B in 45 seconds.

Clearly, A beats B by (45-36)=9 seconds

Speed of B = Distance Time=10045 m/sDistance Covered by B in 9 seconds = Speed × Time = 10045×9 = 20 metre i.e., A beats B by 20 metre

Q23. 24, 60, 120, 210, ?

The pattern is + 36, + 60, + 90,…..i.e. + [6 x (6 + 0)], + [6 x (6 + 4)], + [6 x (6 + 9)],…

So, missing term = 210 + [6 x (6 + 15)] = 210 + 126 = 336.