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🔰Byjus hiring all Batches 🔰
10 laks😱
👉Name of Post :Business Development
👉Qualification : BE/B.Tech
👉Batch : Any Batch
👉Salary : 10 laks
🔴Apply Link : https://byjus.com/apply/
ranks=[0,0,0,0,0]
names=['Malaysia','Australia','Germany','Dubai','France']
flag=0
for i in range(0,5):
if flag==1:
break
for j in range(0,5):
r=input()
if r.isnumeric():
r=int(r)
if r>0 and r<=5:
if r==1:
ranks[j]+=1
else:
flag=1
break
else:
flag=1
break
if flag==1:
print("INVALID INPUT")
else:
comb=list(zip(ranks,names))
comb.sort(key = lambda x:x[0])
top = comb[0][0]
for i in range(0,5):
if top==comb[i][0]:
print(comb[i][1])
import sys
items = {1:50, 2:100, 3:40,4:200,5:300}
user_items =[int(x) for x in input().split() ]
quantity = [int(x) for x in input().split()]
membership = input()
tp = 0.0
k=0
for i in user_items:
if i in items:
tp += items[i] * quantity[k]
k = k+1
else:
print("INVALID INPUT")
sys.exit(0)
if((membership == 'y' or membership == "Y") and tp > 1000):
tp = tp - ((15*tp)/100)
elif(membership == 'y' or membership == "Y"):
tp = tp - ((5*tp)/100)
elif(tp > 1000):
tp = tp - ((10*tp)/100)
print(tp,"INR",sep="")
Bunch of keys
x=[]
while True:
x.append(input())
if x[-1]=='q':
break
x.remove('q')
p=x.count('-')
print("BLANK KEYS:"+str(p))
print("TOTAL KEYS:"+str(len(x)))
print("NUMBER OF LOCKS:"+str(len(x)-p))
l = [0,50,100,40,200,300]
item = list(map(int,input().split()))
quan = list(map(int,input().split()))
c = input()
cost = 0
for i in range(len(item)):
cost = cost + quan[i]*l[item[i]]
if cost>100 and c== 'y':
cost = cost-cost*15/100
if cost>1000 and c!='y':
cost = cost-cost*0.1
if max(item)>5:
print('INVALID INPUT')
else:
print(f'{cost}INR')
#include<bits/stdc++.h>
using namespace std;
int main(){
double price[] = {80.0, 130.0, 100.0, 80.0, 90.0,110.0,120.0, 140.0, 70.0,80.0, 130.0, 160.0, 70.0,60.0,40.0,50.0,30.0,40.0,160.0,150.0};
double total = 0;
int ch, q;
char chc;
do{
cin>>ch>>q;
cout<<"more items? (y/n) \n";
cin>>chc;
if(ch > -1 && ch < 20)
total += price[ch-1] * q;
else {
cout<<"INVALID INPUT";
return 0;
}
} while(chc == 'y');
int t = total;
if(t != total)
cout<<"Total Amount "<<total<<" INR";
else cout<<"Total Amount "<<total<<".0 INR";
return 0;
}
💥 Verbal ability:-
1. The most important thing..
Ans: A B
💥 Numerical Ability:-
1.Train travelling at 79km/hr....
Ans - 7.4
2. If (x+10)%..
Ans: 16
3. Value of..
Ans: 18/35
4. A shopkeeoer...
Ans: 11(1/9)
5. A certain loan..
Ans: 1320
6.if a is 20%...
Ans: 25c:18b
7. Let x be upper limit..
Ans: 10.5
8. The speed of boat..
Ans: 24
9. What is the mean mark..
Ans: 16(2/3)
10. (2.4)^2...
Ans: 0.3&0.5
11. Number of pencils..
Ans- 15
12. A&b can complete..
Ans: 15
13. A certain sum amt..
Ans: 12960
14.Difference between age
Ans: 36
15. In a compatetive exam..
Ans: 26
16. √1+√3/2...
Ans: 9
17. The sum if height...
Ans: 10568
18. Sarita wants to gift..
Ans: 15
💥Numerical Ability:-
1. The marks of 10 student..
Ans: 9.0
2. 10% of voters..
Ans: 20000
3. If (x+10)%...
Ans: 16
4. The standard deviation..
Ans: 16&18
(join helping_hand4u for more)
5. Amita earns..
Ans: loss 3.5%
6. In finding hcf..
Ans: 68
7. If the mean 29 obser..
Ans: 63
8. What is the mean of..
Ans: 14.5
9. Two sarees..
Ans: 150
10. The time taken by boat..
Ans: 6.4
11. In 2018, madhu...
Ans: 19.8
12. 3(4/7)/...
Ans: 15/16
13. Table answer..
Ans: 37.5
14. The ratio of two number..
Ans: 6:5
15. If 5x3y...
Ans: 27
(Join helping_hand4u)
16. What is the sum(in rs)..
Ans: 25020
17. A certain sum..
Ans:15972
18. A metallic speher..
Ans: 185
19. Two soln a and b..
Ans: 45
Join helping_hand4u for more
20. The loan of inr..
Ans: 10140
21. Pipe a and b..
Ans: 32
//python code
t1=0
flag=0
t2=1
t3=0
fib=[]
odd_cnt=0
even_cnt=0
n=input()
if n.isdigit():
n=int(n)
if n<=5 or n>20:
flag=1
else:
for i in range(0,n):
t1=t2
t2=t3
t3=t1+t2
fib.append(t3)
if t3%2==0:
even_cnt+=1
else:
odd_cnt+=1
if flag==1:
print("INVALID INPUT")
else:
for i in range(0,n):
if i!=(n-1):
print(fib[i],end=" ")
else:
print(fib[i])
print(even_cnt)
print(odd_cnt)
//python code
t1=0
flag=0
t2=1
t3=0
fib=[]
odd_cnt=0
even_cnt=0
n=input()
if n.isdigit():
n=int(n)
if n<=5 or n>20:
flag=1
else:
for i in range(0,n):
t1=t2
t2=t3
t3=t1+t2
fib.append(t3)
if t3%2==0:
even_cnt+=1
else:
odd_cnt+=1
if flag==1:
print("INVALID INPUT")
else:
for i in range(0,n):
if i!=(n-1):
print(fib[i],end=" ")
else:
print(fib[i])
print(even_cnt)
print(odd_cnt)
Quants
1. If 3√4913/4096..value of a
A. 1
2. In a competitive exam
A. 26
3. Csa of cone...
A. 11440
4.Two cars a and b meet..
A. 60 kmph
Join @ tcsnqtexam
5. The value 7.6*3.8*15.28..
A. 13 and 14
6. Salary in rupees of 10 emp
A. 11.983
7. Bike..
A. 7
Join @ tcsnqtexam
8. If the perimeter of rectangle...
A. 420
9. Number divisible by 7
A. 2
10. A sum was lent at
A. 3000
Join @ tcsnqtexam
11. A can do a piece..
A. 30 6/7
12. Mean of 29 observations
A. 63
13. Pie Chart
A. 1500
14. if √2 = 1.1414....
A. 1.284
15. A sells an article
A. 690
Join @ tcsnqtexam
16. What percentage of total
A. 52.7
17. A, B and C enter
A. 300000
18. 10% of the voters
A. 25000
Join @ tcsnqtexam
19. If a deaer
A. 1.5
20. A and B undertake
A. 360
Join @placementupdatess
💥 Numerical Ability:-
1.Standard deviation...
Ans: 11.983
2. Mean of the..
Ans: 63
3. If 3√4913/4096...
Ans: 1
4. If the perimeter of rectangle...
Ans: 420
5. Csa of cone...
Ans: 11440
6. The mean of data..
Ans: 5
7. Bike..
Ans: 7
8.Two cars a and b meet..
Ans: 60 kmph
9. A sales representative..
Ans: 50000
10.Valye of 7.6*......
Ans: 13 and 14
11. If the number is divided by 7..
Ans: 4
12. If √2=1.414..
Ans: 1.2843
13. In a compatitive exam..
Ans: 26
14. Pie chart
Ans: 1500
Update by iON,
Students can try resuming back the test and continue the test.
If still the issue persists, no need to panic they will receive the communication accordingly by EOD from iON.
Thank you.
Try Not To Do anytype of Malpractice In TCS nqt...
Even if u do ,do with great care and without being caught
Because TCS may ban your Account and You cannot write NQT after 3 months again...😐
Click on the above link-->Login to the account-->Click here to view updates on your NQT details (Right side of home pages)-->Click on login details tab-->You will able to see slot details
https://learning.tcsionhub.in/iDH/India/home
Reminder!! - Last Date is Approaching soon, Apply Now!!
JOB opportunity with 6.5 LPA Package!!
Cisco Off Campus Drive 2020 | B.E/B.Tech/M.Tech/MCA
Apply Link👇🏻
https://fresheropenings.com/cisco-off-campus-drive-2020-2/
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