**AMCAT**

**AMCAT Topics**

**Updated :**

**8. Eesha bought 18 sharpeners for Rs.100. She paid 1 rupee more for each white sharpener than for each brown sharpener. What is the price of a white sharpener and how many white sharpener did she buy ?**

### 1. 11, 23, 47, 83, 131, . WHAT IS THE NEXT NUMBER?

a. 145

b. 178

c. 176

d. 191

Explanation:

11,23,47,83,131

23–11 = 12

47–23 = 24

83–47 = 36

131–83 = 48

Therefore, 131+60=191

### 2. A SERIES OF BOOK WAS PUBLISHED AT SEVEN YEAR INTERVALS. WHEN THE SEVENTH BOOK WAS PUBLISHED THE TOTAL SUM OF PUBLICATION YEAR WAS 13, 524. FIRST BOOK WAS PUBLISHED IN?

a. 1911

b. 1910

c. 2002

d. 1932

Answer:

Explanation:

Let the years be n, n+7, n+14, …., n+42. (∵∵ use formula Tn=a+(n−1)dTn=a+(n−1)d to find nth term)

Sum = Sn=n2(2a+(n−1)d)Sn=n2(2a+(n−1)d) = 72(2n+(7−1)7)72(2n+(7−1)7) = 13,524

⇒7n+147=13,524⇒7n+147=13,524

⇒⇒ n = 1911

### 3. CRUSOE HATCHED FROM A MYSTERIOUS EGG DISCOVERED BY ANGUS, WAS GROWING AT A FAST PACE THAT ANGUS HAD TO MOVE IT FROM HOME TO THE LAKE. GIVEN THE WEIGHTS OF CRUSOE IN ITS FIRST WEEKS OF BIRTH AS 5, 15, 30,135, 405, 1215, 3645. FIND THE ODD WEIGHT OUT.

a) 3645

b) 135

c) 30

d) 15

Answer: c

Explanation:

5×3 = 15

15×3 = 45 ⇒⇒ Given as 30

45×3 = 135

135×3 = 405

405×3 = 1215

1215×3 = 3645

### 4. A CAN COMPLETE A PIECE OF WORK IN 8 HOURS, B CAN COMPLETE IN 10 HOURS AND C IN 12 HOURS. IF A,B, C START THE WORK TOGETHER BUT A LAVES AFTER 2 HOURS. FIND THE TIME TAKEN BY B AND C TO COMPLETE THE REMAINING WORK.

1) 2 (1/11) hours

2) 4 (1/11) hours

3) 2 (6/11) hours

4) 2 hours

Explanation:

A,B,C’s 1 hour work is = 18+110+11218+110+112 = 15+12+10120=3712015+12+10120=37120

A,B,C worked together for 2 hours, Therefore, 2 hours work is = 37120×2=376037120×2=3760

Remaining work = 1−3760=23601−3760=2360

(23/60 work is done by B and C together)

B, C’s 1 hour work = 110+112=6+560=1160110+112=6+560=1160

(2360)th(2360)th part of the work done by B, C in = (2360)1160(2360)1160 = 21112111 hours.

### 5. A TREE OF HEIGHT 36M IS ON ONE EDGE OF A ROAD BROKE AT A CERTAIN HEIGHT. IT FELL IN SUCH A WAY THAT THE TOP OF THE TREE TOUCHES THE OTHER EDGE OF THE ROAD. IF THE BREADTH OF THE ROAD IS 12M, THEN WHAT IS THE HEIGHT AT WHICH THE TREE BROKE?

a. 16

b. 24

c. 12

d. 18

Explanation:

Let the tree was broken at x meters height from the ground and 36 – x be the length of other part of the tree.

From the diagram, (36−x)2=x2+122(36−x)2=x2+122

⇒1296−72x+x2=x2+144⇒1296−72x+x2=x2+144

⇒72x=1296−144⇒72x=1296−144

⇒x=16⇒x=16

### 6. THE STICKS OF SAME LENGTH ARE USED TO FORM A TRIANGLE AS SHOWN BELOW.IF 87 SUCH STICKS ARE USED THEN HOW MANY TRIANGLES CAN BE FORMED?

Explanation:

First triangle is formed by using 3 sticks, but any subsequent triangle may be formed by using 2 sticks. Therefore, If 1st triangles uses 3 sticks, Remaining sticks = 87 – 3 = 84. With these 84, we can form 42 triangles. So total = 42 + 1 = 43

**Shortcut:**

To solve questions like these, use formula, 2n + 1 = k. Here n = triangles, k = sticks

2n+1 = 87 ⇒⇒ n = 43.

### 7. 17 × 8 M RECTANGULAR GROUND IS SURROUNDED BY 1.5 M WIDTH PATH. DEPTH OF THE PATH IS 12 CM. GRAVEL IS FILLED AND FIND THE QUANTITY OF GRAVEL REQUIRED.

a. 5.5

b. 7.5

c. 6.05

d. 10.08

Explanation:

Area of the rectangular ground = 17×8=136m217×8=136m2

Area of the big rectangle considering the path width = (17+2×1.5)×(8+2×1.5)=220m2(17+2×1.5)×(8+2×1.5)=220m2

Area of the path = 220−136=84m2220−136=84m2

Gravel required = 84m2×12100m=10.08m384m2×12100m=10.08m3

### 8. A SUM OF RS.3000 IS DISTRIBUTED AMONG A, B, AND C. A GETS 2/3 OF WHAT B AND C GOT TOGETHER AND C GETS 1/3 OF WHAT A AND B GOT TOGETHER, C’S SHARE IS?

Explanation:

Let B+C together got 3 units, then A get 2 units. or B+CA=32B+CA=32 – – – (1)

Let A+B together got 3 units, then B get 1 units. or A+BC=31A+BC=31 – – – (2)

By using Componendo and Dividendo, we can re-write equations (1) and (2), A+B+CA=3+22=52=208A+B+CA=3+22=52=208 and A+B+CC=3+11=41=205A+B+CC=3+11=41=205

So A = 8, B = 7, C = 5

C’s share = 5(8+5+7)×3000=7505(8+5+7)×3000=750

### 9. THE NUMBERS 272738 AND 232342, WHEN DIVIDED BY N, A TWO DIGIT NUMBER, LEAVE A REMAINDER OF 13 AND 17 RESPECTIVELY. FIND THE SUM OF THE DIGITS OF N?

a. 7

b. 8

c. 5

d. 4

Explanation:

From the given information, (272738 – 13, 232342 – 17) are exactly divisible by that two digit number.

We have to find the HCF of the given numbers 272725, 232325.

HCF = 25.

So sum of the digits = 7.

### 10. ASSUME THAT F(1)=0 AND F(M+N)=F(M)+F(N)+4(9MN-1). FOR ALL NATURAL NUMBERS (INTEGERS>0)M AND N. WHAT IS THE VALUE OF F(17)?

a. 5436

b. 4831

c. 5508

d. 4832

Explanation:

f(1) = 0

f(2) = f(1+1) = f(1)+f(1)+4(9×1×1 – 1) = 0+0+4×8 = 32

f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204

f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980

f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260