Directions (Q1 - Q2): Study the following information and answer the questions that follow:
The Table given below shows the percentage of valid voters in 5 villages in two years 2001 and 2005. Study the graph carefully to answer the based questions.
NOTE - Total voters in any year = Valid voters + Invalid voters
Q1. What is the difference between invalid voters of village C in the two given years if valid voters in 2005 in that village are 4000 and the ratio of valid voters of village C in 2001 to 2005 is 19: 25?
A. 1190
B. 1250
C. 1290
D. 1350
E. 1365
Q2. If in village A in 2005, 2500 voters were declared invalid voters 10% of valid voters opted NOTA and the winner got 200 more votes than losing candidate, then find the total vote that losing candidate got in 2005 in village A.
A. 4400
B. 4600
C. 5400
D. 5200
E. 4800
Q3. In how many different ways can the letters of the word "COMPLAINT" be arranged in such a way that the vowels occupy only the odd positions?
A. 1440
B. 43200
C. 1440
D. 5420
E. None of these
Q4. In how many different ways can the letters of the word "CANDIDATE" be arranged in such a way that the vowels always come together?
A. 4320
B. 1440
C. 720
D. 840
E. None of these
Q5. 5/9 of 504 + 3/8 of 640 = ?
A. 620
B. 550
C. 520
D. 480
E. 460
Q6. 4900 ÷ 28 * 444 ÷ 12 = ?
A. 6312
B. 6223
C. 6475
D. 6217
E. 6421
Q7. Section A and section B of 7th class in a school contains total 285 students. Which of the following can be a ratio of the ratio of the number of boys and number of girls in the class?
A. 6 : 5
B. 10 : 9
C. 11 : 9
D. 13 : 12
E. Cannot be determined
Q8. There are 50 students in a hostel. Now the number of students got increased by 8. Due to this the expenses of the mess increased by 30 rupees per day while the average expenditure is decreased by 2 rupees. Find the original expenditure.
A. 812.5 rupees
B. 912.5 rupees
C. 1012.5 rupees
D. 1112.5 rupees
E. None of these
Q9. The sum of 3 consecutive even number is 40 more than the average of these numbers. Which of the following is the 2nd largest number ?
A. 18
B. 20
C. 22
D. 24
E. None of these
Q10. What is the least number that can be added to 7218 make it a perfect square ?
A. 7
B. 5
C. 6
D. 4
E. None of these
Q11. A toy is sold for Rs.336, if the % of profit is equal to cost price, then find CP ?
A. 137
B. 140
C. 135
D. 120
E. None of these
Q12. A bought ‘X’ quantity of pencils for Rs.500 and he sold 2/5 part for 5% loss and at which percent he must sell his rest of the quantity so that he would gain overall 10% profit ?
A. 10%
B. 20%
C. 17%
D. 15%
E. None of these
Q13. Dinesh does 80% of a work in 20 days. He then calls in Gokul and theytogether finish the remaining work in 3 days. How long Gokul alone would take to do the whole work?
A. 39 days
B. 37 days
C. 37 [latex]\frac{1}{2}[/latex] days
D. 40 days
E. 39 [latex]\frac{1}{2}[/latex] days
Q14. Hari and Vijay can together finish a work in 30 days. They worked together for 20 days and then Vijay left. After another 20 days, hari finished the remaining work. In how many days hari alone can finish the work?
A. 45
B. 60
C. 35
D. 50
E. 65
Q15. A thief is spotted by a policeman from a distance of 200 metre. When the policeman starts chasing , the thief also starts running. If the speed of the thief be 16kmph and that of policeman be 20kmph, how far the thief will have run before he is overtaken?
A. 800 m
B. 700 m
C. 650 m
D. 750 m
E. None of these
Q16. A bus travels at the rate of 54 kmph without stoppages and it travels at 45 kmph with stoppages. How many minutes does the bus stop on an average per hour?
A. 8 minutes
B. 6 minutes
C. 10 minutes
D. 4 minutes
E. None of these
Q17. Vithya gave Rs.20000/- to Priya for 5 years and Rs.30000/- to Vijay for 2 years. If the rate of interests is the same and she received Rs.12800- as SI from both. What is the rate of interest?
A. 8%
B. 7%
C. 9%
D. 11%
E. None of these
Q18. The difference between CI and SI on the same sum for 2 yrs at the rate of 25% pa is Rs.1500. What is the principal lent out in each case ?
A. Rs.18000
B. Rs.21000
C. Rs.23000
D. Rs.24000
E. None of these
Q19. What should come at the place of question mark (?) in the following number series?
509, 68, 429, 140, 365, ?
A. 194
B. 192
C. 196
D. 195
E. 197
Q20. What should come at the place of question mark (?) in the following number series?
2√3, 2√5, √30, √42, 2√14, ?
A. 6√2
B. √60
C. 5√3
D. 3√5
E. √50
Q21. In a 70 litres mixture of milk and water, % of water is 30%. The milkman gave 20 litres of this mixture to a customer and then added 20 litres of water to the remaining mixture. What is the % of milk in the final mixture ?
A. 48%
B. 50%
C. 40%
D. 53%
E. None of these
Q22. Rani mixes 70kg of sugar worth Rs.28.50 per kg with 100kg of sugar worth Rs.30.50 per kg. At what rate shall he sell the mixture to gain 10% ?
A. Rs.32.64
B. Rs.31.50
C. Rs.30.46
D. Rs.29.75
E. None of these
Q23. The greatest possible length which can be used to measure exactly the lengths 1m 92cm,3m 84cm ,23m 4cm
A. 23
B. 32
C. 36
D. 34
E. None of these
Q24. The HCF of 2511 and 3402 is
A. 31
B. 42
C. 76
D. 81
E. None of these
Q25. I. 3p[latex]^{2}[/latex] + 15p = -18
II. q[latex]^{2}[/latex] + 7q + 12 = 0
A. if P is greater than q.
B. if p is smaller than q.
C. if p is equal to q.
D. if p is either equal to or greater than q.
E. if p is either equal to or smaller than q.
Q26. I. 3x[latex]^{2}[/latex] + 16x + 21 = 0
II. 6y[latex]^{2}[/latex] + 17y + 12 = 0
A. if x < y
B. if x ≤ y
C. relationship between x and y cannot be determined
D. if x ≥ y
E. if x > y
Q27. 84, 72, 66, ?, 68.17
A. 65
B. 60
C. 67
D. 76
E. None of these
Q28. 110, ?, 42, 20, 6
A. 63
B. 81
C. 72
D. 90
E. None of these
Q29. 35% of √3136 x 5 = ? + 54
A. 46
B. 48
C. 44
D. 42
E. None of these
Q30. 90% of 700 + 50% of 1000 – 170 = ?
A. 950
B. 930
C. 960
D. 970
E. None of these
Q31. A said to B – seven years ago I was seven times as old as you were and three year hence I will be three times as old as you will be. The ages of A and B
A. 52, 12
B. 42, 12
C. 32, 12
D. 54, 14
E. None of these
Q32. The ratio of the ages of priya and riya is 2:5. After 8 years, their ages will be in the ratio of 2:3. Then the sum of their present ages
A. 12 years
B. 14 years
C. 16 years
D. 18 years
E. None of these
Q33. The average weight of 40 students in a class is 75 kg. By mistake the weights of two students are read as 74 kg and 66 kg respectively instead of 66 kg and 54 kg. Find the corrected average weight of the class
A. 73.50 kg
B. 74.50 kg
C. 75.50 kg
D. 76.50 kg
E. None of these
Q34. The average weight of 40 balls is 5 grams. When the weight of the basket is added to the weight of balls, the average increased by 0.5 grams. Find the weight of the basket.
A. 20.5 gm
B. 22.5 gm
C. 25.5 gm
D. 28.5 gm
E. None of these
Q35. A deer and a rabbit can complete a full round on a circular track in 9 minutes and 5 minutes respectively. P, Q, R and S are the four consecutive points on the circular track which are equidistant from each other. P is opposite to R and Q is opposite to S. After how many minutes will they meet together for the first time at the starting point, when both have started simultaneously from the same point in same direction?
A. 15 minutes
B. 25 minutes
C. 35 minutes
D. 45 minutes
E. None of these
Q36. A well with 28 m inside diameter is dug out 18 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embarkment. Find the height of the embarkment.
A. 3 m
B. 8 m
C. 9 m
D. 6 m
E. None of these
Q37. A dealer sells a goat for rupees 600 and there by gains 20 percent. He sells another goat at 5 percent loss and on the whole there is no loss no profit. Find the cost price of the second goat
A. 1000
B. 2000
C. 3000
D. 4000
E. None of these
Q38. A toys manufacturing company declares 20 percent discount for whole sale buyers. A trader bought toys for rupees 4000 after getting the discount. Now he sells the toys at 15 percent profit on the original price of the company. What is the selling price of the toys?
A. 4750
B. 5740
C. 5750
D. 6750
E. None of these
Q39. 19, 23, 14, 30, 5, ?
A. 45
B. 41
C. 51
D. 35
E. None of these
Q40. 12, 8, 16, 48, 184, ?
A. 560
B. 420
C. 860
D. 550
E. None of these
Answers and Explanations
1. Answer - Option C
Explanation -
Invalid voter of village C in 2005 = 4000 * [latex]\frac{36}{64}[/latex] = 2250
Valid voter of village C in 2001 = 4000 * [latex]\frac{19}{25}[/latex] = 3040
Now, invalid voter of village C in 2001 = 3040 * [latex]\frac{24}{76}[/latex] = 960
∴Required diff. = 2250 – 960 = 1290
2. Answer - Option A
Explanation -
Total valid votes of village A in 2005 = 2500 * [latex]\frac{100}{20}[/latex] * [latex]\frac{80}{100}[/latex] = 10000
Total valid votes excluding Nota in village A in 2005 = 10000 * [latex]\frac{90}{100}[/latex] = 9000
∴According to question,
x + (x + 200) = 9000
--> x = 4400
Required no. of votes of losing candidates = 4400
3. Answer - Option B
Explanation -
There are 9 different letters in the given word "COMPLAINT", out of which there are 3 vowels and 6 consonants.
Let us mark these positions as under:
[1] [2] [3] [4] [5] [6] [7] [8] [9]
Now, 3 vowels can be placed at any of the three places out of 5 marked 1, 3, 5, 7 and 9.
Number of ways of arranging the vowels = 5P3 = 5x4x3 = 60 ways.
Also, the 6 consonants at the remaining positions may be arranged in 6P6 ways = 6! ways = 720 ways.
Therefore, required number of ways = 60 x 720 = 43200 ways.
4. Answer - Option A
Explanation -
There are 9 letters in the given word, out of which 4 are vowels.
In the word "CANDIDATE" we treat the vowels "AIAE" as one letter.
Thus, we have CNDDT(AIAE).
Now, we have to arrange 6 letters, out of which D occurs twice.
Therefore, number of ways of arranging these letters = [latex]\frac{6!}{2!}[/latex] = [latex]\frac{720}{2}[/latex] = 360 ways.
Now, AIAE has 4 letters, in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = [latex]\frac{4!}{2!}[/latex] = [latex]\frac{1 * 2 * 3 * 4}{2}[/latex] = 12
Therefore, required number of words = (360 x 12) = 4320.
5. Answer - Option C
Explanation -
[latex]\frac{5}{9}[/latex] of 504 = 280 ; [latex]\frac{3}{8}[/latex] of 640 = 240 => 280 + 240 = 520
6. Answer - Option C
Explanation -
4900 ÷ 28 = 175
444 ÷ 12 = 37
175 * 37 = 6475
7. Answer - Option B
Explanation -
The number of boys and girls cannot be in decimal values, so the denominator should completely divide number of students (285).
Check each option:
6 + 5 = 11, and 11 does not divide 285 completely.
10 + 9 = 19, and only 19 divides 285 completely among all.
8. Answer - Option B
Explanation -
Let initial expenditure is E per day. Now it is increased by 30 rupees per day,
Initial students = 50 and now they are 58,
[latex]\frac{E}{50}[/latex] – [latex]\frac{(E + 30)}{58}[/latex] = 2
Solve for E, We will get E = 912.5 rupee.
9. Answer - Option B
Explanation - [latex]\frac{(x + x + 2 + x + 4)}{3}[/latex]
X + x + 2 + x + 4 = 40 + [latex]\frac{(x + x + 2 + x + 4)}{3}[/latex]
3x + 6 = 40 + [latex]\frac{(3x+6)}{3}[/latex]
[latex]\frac{(9x + 18 – 3x - 6)}{3}[/latex] = 40
6x + 12 = 120
X = [latex]\frac{108}{6}[/latex] = 18
18 + 2 = 20
10. Answer - Option A
Explanation -
85 * 85 = 7225
11. Answer - Option B
Explanation -
(100 + x)% of X = 336
100x + x[latex]^{2}[/latex] = 33600
x[latex]^{2}[/latex] + 100x – 33600 = 0
(x + 240x)(x - 140x) = 0
X = 140
12. Answer - Option B
Explanation -
500 * [latex]\frac{2}{5}[/latex] = 200
5%L = 200 * [latex]\frac{95}{100}[/latex] = 190
500 + 10 + ([latex]\frac{500 * 10}{100}[/latex]) = 510 + 50 = 560
[latex]\frac{60}{500}[/latex] * 100 = 20%
13. Answer - Option C
Explanation -
Dinesh work done = 20 * [latex]\frac{5}{4}[/latex] = 25 days
[latex]\frac{1}{5}[/latex] workdone by Dinesh and gokul in 3days.
Whole work done = 15 days
Dinesh 1 days work = [latex]\frac{1}{25}[/latex]
Dinesh and gokul’s 1 day work = [latex]\frac{1}{15}[/latex]
Gokul’s 1 day work = [latex]\frac{1}{15}[/latex] – [latex]\frac{1}{25}[/latex] = [latex]\frac{2}{75}[/latex]
Gokul alone in [latex]\frac{75}{2}[/latex] days or 37 [latex]\frac{1}{2}[/latex] days.
14. Answer - Option B
Explanation -
Hari + vijay 20 days work = [latex]\frac{1}{30}[/latex] * 20 = [latex]\frac{2}{3}[/latex]
Remaining work = [latex]\frac{1}{3}[/latex]
[latex]\frac{1}{3}[/latex] work in 20 days so whole work in 60 days.
15. Answer - Option A
Explanation -
d = 200 m, a = 16kmph = [latex]\frac{40}{9}[/latex] m/s, b = 20kmph = [latex]\frac{50}{9}[/latex] m/s
Required Distance D = d * ([latex]\frac{a}{b-a}[/latex])b= 200 * ([latex]\frac{40}{9}[/latex] * [latex]\frac{9}{10}[/latex]) = 800m
16. Answer - Option C
Explanation -
Due to stoppages, the bus can cover 9 km less per hour[54 - 45 = 9] Time taken to cover 9 km = ([latex]\frac{9}{54}[/latex]) x 60 = 10 minutes.
17. Answer - Option A
Explanation -
20000 * 5 * [latex]\frac{R}{100}[/latex] + 30000 * 2 * [latex]\frac{R}{100}[/latex] = 12800
1600R = 12800
R = [latex]\frac{12800}{1600}[/latex] = 8
18. Answer - Option D
Explanation -
P[latex](\frac{r}{100})^{2}[/latex] = 1500
P[latex](\frac{25}{100})^{2}[/latex] = 1500
P * [latex]\frac{1}{16}[/latex] = 1500
P = 1500 * 16 = 24000
19. Answer - Option C
Explanation -
20. Answer - Option E
Explanation -
21. Answer - Option B
Explanation -
20litre given remaining = 70 - 20 = 50litre
Quantity of milk = 50 * [latex]\frac{70}{100}[/latex] = 35litre
Quantity of water = 50 - 35 = 15litre
20litres of water added = 50 + 20 = 70
% of milk = 35 * [latex]\frac{100}{70}[/latex] = 50%
22. Answer - Option A
Explanation -
[latex]\frac{70}{100}[/latex] = 100x – [latex]\frac{30.50(110)}{28.50(110)}[/latex] - 100x
0.7 = 100x - [latex]\frac{3355}{3135}[/latex] - 100x
2194.5 – 70x = 100x - 3355
170x = 5549.5
X = 32.64
23. Answer - Option B
Explanation -
192 = 4 × 2 × 2 × 2 × 3
384 = 4 × 2 × 2 × 2 × 6
2304 = 4 × 2 × 2 × 6 × 2
HCF = 4 × 2 × 2 = 16 × 2 = 32
24. Answer - Option D
Explanation -
2511 = 81 × 31
3402 = 81 × 42
Hence HCF is 81
25. Answer - Option D
Explanation -
I. 3p[latex]^{2}[/latex] + 15p = -18
[latex] \rightarrow[/latex] p[latex]^{2}[/latex] + 5p + 6 =0
[latex] \rightarrow[/latex] p = -2, -3
II. q[latex]^{2}[/latex] + 7q + 12 = 0
[latex] \rightarrow[/latex] (q + 3) (q + 4) = 0
[latex] \rightarrow[/latex] q = -3, -4
26. Answer - Option A
Explanation -
I. 3x[latex]^{2}[/latex] + 16x + 21 = 0
[latex] \rightarrow[/latex] 3x[latex]^{2}[/latex] + 9x + 7x + 21 = 0
[latex] \rightarrow[/latex] (x + 3) (3x + 7) = 0
[latex] \rightarrow[/latex] x = -3, -[latex]\frac{7}{3}[/latex]
II. 6y[latex]^{2}[/latex] + 17y + 12 = 0
[latex] \rightarrow[/latex] 6y[latex]^{2}[/latex] + 9y + 8y + 12 = 0
[latex] \rightarrow[/latex] 3y (2y + 3) + 4 (2y + 3) = 0
[latex] \rightarrow[/latex] y = -[latex]\frac{3}{2}[/latex], -[latex]\frac{4}{3}[/latex]
27. Answer - Option A
Explanation -
[latex]\frac{84}{1.2}[/latex] = 70 + 2 = 72
[latex]\frac{72}{1.2}[/latex] = 60 + 6 = 66
[latex]\frac{66}{1.2}[/latex] = 55 + 10 = 65
[latex]\frac{65}{1.2}[/latex] = 54.17 + 14 = 68.17
28. Answer - Option C
Explanation -
11[latex]^{2}[/latex] – 11 = 110
9[latex]^{2}[/latex] – 9 = 72
7[latex]^{2}[/latex] – 7 = 42
5[latex]^{2}[/latex] – 5 = 20
3[latex]^{2}[/latex] – 3 = 6
29. Answer - Option C
Explanation -
([latex]\frac{65}{100}[/latex]) * 56 * 5 = ? + 54
98 – 54 = 44
30. Answer - Option C
Explanation -
= (700 * ([latex]\frac{90}{100}[/latex]) + 1000 * ([latex]\frac{50}{100}[/latex])) – 170
= 630 + 500 – 170 = 960
31. Answer - Option B
Explanation -
A – 7 = 7 * (B – 7)....(1)
A + 3 = 3 * (B + 3)....(2)
Solve both equations to get A and B
32. Answer - Option B
Explanation -
Let priya age = 2x and riya age = 5x.
So, [latex]\frac{(2x + 8)}{(5x + 8)}[/latex] = [latex]\frac{2}{3}[/latex], we get x = 2. So sum of priya and riya = 14
33. Answer - Option B
Explanation -
Weight of 40 students = 40 * 75
new weight = 40*75 – 74 – 66 + 66 + 54 = 40*75 – 20
so new average = [latex]\frac{(40*75 – 20)}{40}[/latex] = 74.50 kg
34. Answer - Option C
Explanation -
[latex]\frac{(40*5 + B)}{41}[/latex] = 5.5 (B is the weight of basket)
35. Answer - Option D
Explanation -
Time taken by a deer to complete one round = 9 minutes
Time taken by a rabbit to complete one round = 5 minutes
They meet together for the first time at the starting point = LCM of 9 and 5 = 45 minutes.
36. Answer - Option A
Explanation -
[latex]\frac{22}{7}[/latex][(R[latex]^{2}[/latex]) – (r[latex]^{2}[/latex])] * h = [latex]\frac{22}{7}[/latex](7 * 7 * 18)
[(35[latex]^{2}[/latex]) – (7[latex]^{2}[/latex])]h = 14 * 14 * 18
(42 * 28)h = 14 * 14 * 18
h = 3 m
37. Answer - Option B
Explanation -
600 = ([latex]\frac{120}{100}[/latex]) * cp
so cp of first goat = 500. So in the first deal he gains 100 rupees.
But on selling second horse he losses 5 percent, means ([latex]\frac{5}{100}[/latex]) * CP = 100
So cp of second goat = 2000
38. Answer - Option C
Explanation -
Let original price of toys be M,
([latex]\frac{80}{100}[/latex]) * M = 4000, M = 5000
Now to gain 15% on the original price, the trader will
sell the toys at ([latex]\frac{115}{100}[/latex]) * 5000 = 5750 rupees
39. Answer - Option B
Explanation -
19 + 2[latex]^{2}[/latex] = 23
23 – 3[latex]^{2}[/latex] = 14
14 + 4[latex]^{2}[/latex] = 30....
40. Answer - Option C
Explanation -
12 * 0.5 + 2 = 8
8 * 1.5 + 4 = 16
16 * 2.5 + 8 = 48....