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# Gauss Contest (NMTC PRIMARY 2018 - V and VI Grades) - Stage I- Problems and Solution

## Part A

###### Problem 1

Observe the following sequence. What is the 100th term?
$$7,8,1,0,0,1,0,1,1,0,2,1,0,3, \ldots \ldots \ldots$$
(A) 1
(B) 0
(C) 2
(D) 3

###### Problem 2

A number is multiplied by 2 then by $\frac{1}{3}$, then by 4 , then by $\frac{1}{5}$ then by 6 and finally by $\frac{1}{7}$. The answer is 16 . Then the number is
(A) odd
(B) even
(C) Square
(D) a cube

###### Problem 3

Samrud bought a t- shirt for Rs.250. His friend Shlok wanted by buy it. Samrud wants to have a $10 \%$ profit on that. The selling price is (in rupees)
(A) 280
(B) 278
(C) 276
(D) 275

###### Problem 4

The value of $1+21+4161+81-11-31-51-71-91$ is
(A) -50
(B) 50
(C) 100
(D) -100

###### Problem 5

In the adjoining figure what portion of the figure is shaded ?

(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) $\frac{3}{10}$

###### Problem 6

The sum of the numbers in the three brackets ( ) is
$$\frac{()}{24}=\frac{20}{()}=\frac{24}{18}=\frac{4}{()}$$
(A) 60
(B) 55
(C) 50
(D) 45

###### Problem 7

A is the smallest three digit number which leaves a remainder 2 when divided by $17 . B$ is the smallest three digit number which leaves remainder 7 . When divided by 12 . Then $A+B$ is
(A) 205
(B) 312
(C) 215
(D) 207

###### Problem 8

A square of side 3 cm in cut into 9 equal squares. Another square of side 4 cm is cut into 16 equal squares. Saket made a bigger square using all the smaller square bits. The length of the side of the bigger square is (in cm)

(A) 7
(B) 6
(C) 5
(D) 8

###### Problem 9

A contractor constructed a big hall, rectangular in shape, with length 32 meters and breadth 18 meters. He wanted to buy 1 meter by 1 meter tiles. But in the shop 3 meter by 2 meter tiles only were available. How many tiles he has to buy for tilting the floor?

(A) 48
(B) 96
(C) 120
(D) 126

###### Problem 10

The fraction to be added to $2 \frac{1}{3}$ to get the fraction $4 \frac{4}{7}$ is
(A) $2 \frac{1}{21}$
(B) $2 \frac{4}{21}$
(C) $2 \frac{5}{21}$
(D) $2 \frac{6}{21}$

#### Part B

###### Problem 11

In the adjoining figure $\angle \mathrm{BAD}=\angle \mathrm{DAF}=\angle \mathrm{FAC}$. GE is parallel to $\mathrm{DF}$, and $\angle \mathrm{EGA}=90^{\circ}$. If $\angle \mathrm{ACE}=70^{\circ}$, the measure of $\angle \mathrm{FDE}$ is $\rule{1cm}{0.15mm}$.

###### Problem 12

$A B C$ is a triangle in which the angles are in the ratio $3: 4: 5$. PQR is a triangle in which the angles are in the ratio $5: 6: 7$. The difference between the least angle of $A B C$ and the least angle of PQR is $a^{\circ}$. Then $a=$ $\rule{1cm}{0.15mm}$.

###### Problem 13

Samrud had to multiply a number by 35 . By mistake he multiplied by 53 and got a result 720 more. The new product is $\rule{1cm}{0.15mm}$.

###### Problem 14

Vishva plays football every 4 th day. He played on a Tuesday. He plays football on a Tuesday again in is $\rule{1cm}{0.15mm}$ days.

###### Problem 15

In an elementary school $26 \%$ of the students are girls. If there are 240 less girls than boys, then the strength of the school is $\rule{1cm}{0.15mm}$.

###### Problem 16

There are three concentric circles as shown in the figure. The radii of them are $2 \mathrm{~cm}, 4 \mathrm{~cm}$ and 6 $\mathrm{cm}$. The ratio of the area of the shaded region to the area of the dotted region is $\frac{a}{b}$ where $a, b$ are integers and have no common factor other than 1. Then $a+b=$ $\rule{1cm}{0.15mm}$.

###### Problem 17

The value of $\left(1+\frac{1}{9}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{7}\right)\left(1+\frac{1}{6}\right)\left(1+\frac{1}{5}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{2}\right)$ is $\rule{1cm}{0.15mm}$.

###### Problem 18

When a two digit number divides 265 , the remainder is 5 . The number of such two digit numbers is $\rule{1cm}{0.15mm}$.

###### Problem 19

If $A \# B=\frac{A \times B}{A+B}$, the value of $\frac{12 \# 8}{8 \# 4}+\frac{10 \# 6}{6 \# 2}$ is $\rule{1cm}{0.15mm}$.

###### Problem 20

When water becomes ice, its volume increases by $10 \%$. When ice melts into water its volume decreases by $a \%$. Then $a=$ $\rule{1cm}{0.15mm}$.

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