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Direction: What will come in place of the question mark (?) in the following number series?
1342, 2210, ?, 4930, 6878
21, 45, 78, 122, ?, 251
5, 7, 13, 23, 41, ?
12, 16, 36, 136, 636, ?
14, 70, 490, 6370, ?
Directions: The following graph represents the number of students studying three different subjects individually in four colleges. Study the graph carefully and answer the questions given below.
What is the ratio of the number of students studying math in college B to that studying English in college C?
180:75 =12:5
25% of total number of students studying math in all college together is what percent more than that of 20% of total number of students studying Science in all college together?
Total number of student studying maths= 690 Totla numbe of students studying scienc = 450 25% of 690= 172.5 20% of 450= 90 so, required percentage= [(172.5-90)*100]/90 = 825/9 = 91.66
Number of students studying English in college D is what percent of number of students studying Science in college A
Required percentage = 180*100/140 = 128.57
Find the difference between number of students studying math in college C and D together and number of students studying Science in college A and B together
Required difference = (200 +160) –(140 +90) =360 -230 =130
Find the ratio between 40% of total number of students studying math in all colleges together to 30% of total number of students studying English in all colleges together?
=184:101
Directions: In each of the following question, a pair of equations is given. You have to find out the values of X and Y and give answer: I. 4X2 - 4X - 3 = 0 II. 4Y2 + 12Y + 5 = 0
I. 4X2 - 4X - 3 = 0 4X2 - 6X + 2X - 3 = 0 2X(2X - 3) + 1(2X - 3) = 0 X = +3/2, -1/2 II. 4Y2 + 12Y + 5 = 0 4Y2+ 10Y + 2Y + 5 = 0 2Y(2Y + 5) + 1(2Y + 5) = 0 Y = -1/2, -5/2 Hence, X ≥ Y
Directions: In each of the following question, a pair of equations is given. You have to find out the values of X and Y and give answer: I. 4X2 = 49 II. 9Y2 - 66Y + 121 = 0
I.4X2 = 49 X2 = 49/4 X = +7/2, -7/2 II. 9Y2 - 66Y + 121 = 0 9Y2 - 33Y - 33Y + 121 = 0 3Y(3Y - 11) - 11(3Y - 11) = 0 Y = +11/3, +11/3 Hence X < Y
Directions: In each of the following question, a pair of equations is given. You have to find out the values of X and Y and give answer:
I. X2 + 9X + 14 = 0 II. Y2 + Y = 2
I. X2 + 9X + 14 = 0 X2+ 7X + 2X + 14 = 0 X(X + 7) + 2(X + 7) = 0 X = -2, -7 II. Y2 + Y = 2 Y2 + 2Y - Y - 2 = 0 Y(Y + 2) - 1(Y + 2) = 0 Y = +1, -2 Hence, X ≤ Y
Directions: In each of the following question, a pair of equations is given. You have to find out the values of X and Y and give answer: I. 9X2 + 5 = 18X II. 2Y2 - 9Y + 10 = 0
I. 9X2 + 5 = 18X 9X2 - 15X - 3X + 5 = 0 3X(3X - 5) - 1(3X - 5) = 0 X = +1/3, +5/3 II. 2Y2 - 9Y + 10 = 0 2Y2 - 5Y - 4Y + 10 = 0 Y(2Y - 5) - 2(2Y - 5) = 0 Y = +2, +5/2 Hence, X < Y
Directions: In each of the following question, a pair of equations is given. You have to find out the values of X and Y and give answer: I. 2X2 + 7X + 6 = 0 II. 2Y2 + 7Y + 5 = 0
I. 2X2 + 7X + 6 = 0 2X2 + 4X + 3X + 6 = 0 2X(X + 2) + 3(X + 2) = 0 X = -2, -3/2 II. 2Y2 + 7Y + 5 = 0 2Y2 + 2Y + 5Y + 5 = 0 2Y(Y + 1) + 5(Y + 1) = 0 Y = -1, -5/2 Hence, no relation is established
Direction: In the following pie-chart the percentage distribution of the population of six cities is given. Total population of these cities is 36 lakh. The given table shows the ratio of male to female and the percentage of literate population in these cities.
What is the total male population of City S?
Total male population of S = (16/100) * 36* (4/9) = 2.56 lakh
If 25% of male population of City Q is shifted to City R then what is the difference between the male population of City R and the male population of City P?
25% of male population of City Q = (25/100) * (18/100) * 36 * (⅝) = 81/80 Male population of R including 25% from Q = (24/100) * 36 *(⅓) + 81/80 - (a) Male population of P = (13/100) * 36 * (5/13) - (b) = (a) - (b) = 2.09
What is the number of females in City T who are literate?
In the given table percentage of literate is given but the ratio of male to female literate is not given. So we can’t find the number.
The total females present in City T is approximately what percent of the total number of persons in City U who are literate?
Total females in T = (17/100) * 36 * (8/17) = 2.88 Literate people in U = (12/100) * 36 * (45/100) = 1.944 = 2.88/1.944 * 100 = 148%
The number of literate persons in City Q is approximately what percent of the number of males in City S?
Literate in Q = (62/100) * 36 * (18/100) = 4.0176 Male in S = (16/100) * 36 * (4/9) = 2.56 = 4.0176/2.56 * 100 = 156.9 = 157%
Direction: What value should come in place of question mark (?) in the following question?
Directions : Study the following table to answer the given questions. Centre wise and Subjects wise number of Teachers
In Bareli, number of General Science teachers is approximately what per cent of Maths teachers?
Required Percentage = 1800/15900×100 = 11.32 = 11% (approx)
What is the difference between total number of Math’s and Reasoning teachers?
Total no of maths teachers = (3000+25000+27000+4500+15900+12360+10000)= 97760 Total no of reasoning teachers = (6000+18000+20500+21000+18650+16300+12000) = 112450 Required Difference = (112450-97760) = 14690
In Jhansi, the number of Reasoning teachers is how much per cent more than that of maths teachers?
Numbers of reasoning teachers in jhansi = 12000 Numbers of maths teachers in jhansi= 10000 Required percentage ={(12000-10000)/10000}×100 = 20%
Which center has 250% more number of reasoning teachers as compared to those in Patna?
250% means 3.5 times the number reasoning teachers in Patna, No. of reasoning teachers in patna = 6000 250% more = 3.5*6000 = 21000 No of reasoning teachers in Delhi = 21000
Which center has the highest number of teachers?
Total no. of teachers in Patna = (3000+6000+150+2500+1750) = 13400 Total no. of teachers in Gaya = (25000+18000+260+12000+1850) = 57110 Total no. of teachers Noida = (27000+20500+370+8000+2100) = 57970 Total no. of teachers Delhi = (4500+21000+450+10000+2150) = 38100 Total no. of teachers Bareli = (15900+18650+275+2300+1800) = 38925 Total no. of teachers Ranchi = (12360+16300+145+2500+1650) = 32955 Total no. of teachers Jhansi = (10000+12000+250+2650+1500) = 26400 So Noida has highest number of teachers.
If pipes X and Y can fill a tank in 14 minutes and 20 minutes respectively. Both are kept open for 3 minutes and then Y is closed and X fills the rest of the tank in p minutes. For how long was pipe X kept open in all?
7 minutes
Given, pipe X fills the tank in 14 minutes and pipe Y in 20 minutes. Therefore, in 1 minute: X fills 1/14 of the tank and Y fills 1/20 of the tank. When X and Y are kept open for 3 minutes, total tank filled = ⇒1/14 × 3 + 1/20 ×3 ⇒(30+21)/140 ⇒51/140 of the tank ∴Amount of tank left to be filled = 1 - 51/140 = 89/140 ∴Time pipe X will take to fill 89/140th of the tank = 89/140 ÷Amount of tank filled by pipe X in 1 minute = 89/140 ÷ 1/14 = 89/140 ×14 = 8.9 minutes ~ 9 minutes ∴Total time for which pipe X was open = 3 minutes + 9 minutes = 12 minutes
A fruit seller makes 25% profit when he gives 10% discount while selling mangoes. At the end of the day he gives a customer 5% additional discount to sell the remaining mangoes quickly. What is his new percentage of profit (approximately)?
Assume that the marked price of one mango = ₹100 He gives 10% discount while selling mangoes. ⇒Selling price of one mango = 100 – 10 = ₹90 He makes 25% profit. Cost price = ₹90 ÷ 1.25 = ₹72 At the end of the day he gives a customer 5% additional discount. ⇒Net discount = (10 + 5) = 15% When he gives 15% discount then his selling price is: 100 – 15 = ₹85 Now his profit is: ₹85 - ₹72 = ₹13 His new percentage of profit =
200 workers are working in a project to complete it in 100 days. In 75 days it was found out that only half of the work is finished. How many more workers does the team need to complete the project on time?
Work completed in 75 days = 200 *75 = half work = w/2 for rest half work to be done on time i.e. in left 25 days, ' x' men are added 200*75/1/2 = (200+x)25/1/2 600= 200+x x= 400 hence 400 worker are required to complete the work on time
A and B started a business together, A invested 25% more than B and C joined the business with a capital of 20% more than A. They decided to give donation 10% of yearly profit in a trust. If after one year B received Rs 4500 in profit, then find the profit earned by C
Let B invested money Rs B And A invested money =1.25B and C invested the money=1.2 * 1.25b =1.5B Ratio of their profit share =1.25: 1: 1.5 =5: 4: 6 4/15 * 90/100 * t = 4500 Profit share of C =6/15*(0.9*18750) = 6750
A man travels a distance with a constant speed. If he travels with 1 km/hr higher speed he will take 4 hours less to travel same distance and if he travels with 1km/hr lower speed, he will take 6 hours more to travel same distance find the distance, travelled by him.
A square is formed from a wire of certain length whose area is 169 sq cm. A rectangle is formed from the same wire. What will be the length of rectangle if breadth of rectangle is 60% of its length?
Side of square
cm Perimeter of square = 4 *13 = 52 cm Let the length of rectangle is x and breadth is 60% of x Thus breadth =3x/5 Perimeter of rectangle = 2( l+ b)
The difference between simple and compound interest at rate of 4% for 2 years is Rs. 24. If the interest is compounded half yearly, what is the difference between simple and complex interest for 1 years?
Given, Let the principal amount be Rs. x. In case of simple interest, S.I = PTR/100 [Where, P = Principal amount, T = duration in years, R = rate of interest] = x × 2 × 4% = 2x/25 In case of compound interest, C.I.
=
[Where, P = Principal amount, r = interest percentage, n = duration in years] = = x × (51/625) Now, we can write, (51x/625) – (2x/25) = 24 ⇒ x/625 = 24 ⇒ x = 15000 So, the simple interest of 1 year = Rs. 15000 × 1 × 4% = Rs. 600. If the interest is compounded half yearly, C.I for 1 year, C.I. = = = = 15000 × (101/2500) = 606 ∴ The difference between simple and complex interest for 1 years, when the interest is compounded half yearly = Rs. (606 – 600) = Rs. 6.
A and B together can do a piece of work in 15 days and B and C can do the same work in 20 days. C alone can do the same work in 30 days. A and C start the work together and do the work for 5 days after that B joined them. In how many days the work will be completed?
Work done by A and B in a day = 1/15 ....eqn (1) Work done by B and C in a day = 1/20 ....eqn (2) Work done by c in a day = 1/30 ..............eqn (3) One day work of B = eqn(2) - eqn(3) = 1/20 - 1/30 = 1/60..eqn(4) One day work of A = eqn(1) - eqn(4) = 1/15 - 1/60 = 1/20 One day work of A and C= 1/20 + 1/30 = 5/60 = 1/12 5 day work of A and C = 5/12 Remaining work = 1- 5/12 = 7/12 One day work of A, B and C = 1/15 + 1/30 = 1/10 Remaining work done by A, B and C together = 7/12 * 10 = 35/6 = 35/6 Total days = 5 + 35/6 = 65/6
From a bag containing 6 green balls and 4 red balls, 3 balls are drawn one after the other. Find the probability of all three balls being green if the balls drawn are not replaced?
Probability of first ball being green = (6/10) Probability of second ball being green = (5/9) [Now number of green balls left are 5 & total balls are 9] Probability of third ball being green = (4/8) Hence, Probability
= 69/10 * 5/9 * 4/8 = 1/6
A shopkeeper buys 200 articles at the rate of Rs 4 per article and he spendsof the cost price at transport. If he sold all article at the rate of Rs 75 per dozen, then find his profit percentage.
Cost price Cost price of one article = Selling price of one article Profit percentage
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